初二电学经典题目。（越多越好） 谁有经典的电学题目，是初二的。
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A\uff0e15W\u7684\u706f\u6ce1\u70e7\u574f\uff0c100W\u7684\u706f\u6ce1\u5b8c\u597d B\uff0e100W\u7684\u706f\u6ce1\u70e7\u574f\uff0c15W\u7684\u706f\u6ce1\u5b8c\u597d
C\uff0e\u4e8c\u53ea\u706f\u6ce1\u5747\u88ab\u70e7\u574f D\uff0e\u4e8c\u53ea\u706f\u6ce1\u5747\u5b8c\u597d
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A\uff0e\u56db\u706f\u4e32\u8054 B\uff0e\u56db\u706f\u5e76\u8054
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12\u4f0f\uff0c \u5219\u7535\u8def\u603b\u7535\u963b\u4e3a________\uff0c\u7535\u538b\u8868\u793a\u6570\u4e3a__________\uff0c\u7535\u6d41\u8868\u793a\u6570
\u4e3a__________\u3002
11\uff0e\u201c220V15W\u201d\u7684\u767d\u70bd\u706f\uff0c\u5176\u706f\u4e1d\u957f\u4e3a5\u5398\u7c73\uff0c\u65ad\u4e86\u540e\u518d\u5854\u63a5\u4e0a\uff0c\u5176\u957f\u5ea6\u7f29\u77ed\u4e861\u5398\u7c73\uff0c\u82e5\u4e0d\u8003\u8651\u642d\u63a5\u5904\u706f\u4e1d\u7c97\u7ec6\u7684\u53d8\u5316\u53ca\u6e29\u5ea6\u5bf9\u7535\u963b\u7684\u5f71\u54cd\uff0c\u5219\u5c06\u5b83\u63a5\u5728220\u4f0f\u7684\u7535\u8def\u4e0a\u4f7f\u7528\u65f6\uff0c\u6d88\u8017\u7684\u529f\u7387\u662f\u539f\u6765\u7684________\u500d\u3002
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13\uff0e\u628a\u6807\u6709\u201cPZ220\u201425\u201d\u548c\u201cPZ220\u201440\u201d\u7684\u4e24\u53ea\u706f\u6ce1\u4e32\u8054\u8d77\u6765\u63a5\u5728
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\u7684\u7535\u8def\u4e2d\uff0c\u8f83\u4eae\u7684\u706f\u662f______\u3002
14\uff0e\u5f00\u5173\u7535\u706f\u65f6\u6536\u97f3\u673a\u91cc\u4f1a\u51fa\u73b0\u201c\u5580\u5566\u201d\u58f0\uff0c\u8fd9\u662f\u56e0\u4e3a\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f\uff3f
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\u7b54\u6848
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\u4e8c\uff0e1\u300120V\uff0c1.25A 2\u3001\u5f02\uff0c\u6469\u64e6 3\u3001\u88c2\u53d8\uff0c\u805a\u53d8\uff0c\u88c2\u53d8 4\u300110,0\uff0c10,5 5\u3001\u8fde\u7ebf\u7565\u3002\u8fd9\u4e2a\u63d2\u5b54\u5177\u6709\u5b89\u5168\u4fdd\u969c\u4f5c\u7528\u3002\u5982\u679c\u5b83\u4e0e\u5927\u5730\u7684\u8fde\u7ebf\u65ad\u4e86\uff0c\u4e00\u65e6\u7535\u5668\u7684\u5916\u58f3\u4e0e\u7535\u6e90\u706b\u7ebf\u4e4b\u95f4\u7684\u7edd\u7f18\u635f\u574f\uff0c\u5bb9\u6613\u9020\u6210\u89e6\u7535\u4e8b\u6545\u30026\u30016,3 7\u30012,2/3 8\u300120\uff0c0.12 9\u3001125 10\u30015\u6b27 9. 6\u4f0f 1. 6\u5b89 11\u30011.25 12\u3001\u51c6\u786e 13\u3001PZ220\u201425\uff0cPZ220\u201440 14\u3001\u5f00\u5173\u7535\u706f\u65f6\uff0c\u7535\u6d41\u53d1\u751f\u8fc5\u901f\u53d8\u5316\uff0c\u4ea7\u751f\u7535\u78c1\u6ce2\uff0c\u6536\u97f3\u673a\u63a5\u6536\u5230\u4e86\u8fd9\u4e00\u7535\u78c1\u6ce2\u5e76\u5c06\u5176\u8f6c\u53d8\u6210\u58f0\u97f3\uff0c\u8fd9\u5c31\u662f\u201c\u5580\u5566\u201d\u58f0\u3002
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\u4ea4\u6d41\u7535\u538b\u8868\uff08\u91cf\u7a0b\u4e3a250V\uff09\u3002
2\uff0e\u6b65\u9aa4\uff1a(1)\u7528\u91cf\u676f\u5411\u58f6\u4e2d\u6ce8\u6c34\uff0c\u91cf\u676f\u5bb9\u5668\u4e58\u6ce8\u6c34\u6b21\u6570\u5f97\u6c34\u7684\u4f53\u79efV\uff1b
(2)\u7528\u6e29\u5ea6\u8ba1\u6d4b\u51fa\u6c34\u7684\u521d\u6e29t0\uff1b(3)\u63a5\u901a\u7535\u6e90\u5e76\u5f00\u59cb\u8ba1\u65f6\uff1b
(4)\u6d4b\u5f97\u6d47\u6c34\u65f6\u7535\u6e90\u7684\u5b9e\u9645\u7535\u538bu\u2019\uff1b
(5)\u6c34\u521a\u6cb8\u817e\u65f6\uff0c\u8ba1\u65f6\u7ed3\u675f\uff0c\u6d4b\u5f97\u52a0\u70ed\u7684\u65f6\u95f4t\uff0c\u7136\u540e\u6d4b\u5f97\u6c34\u6cb8\u817e\u7684\u6e29\u5ea6t\u3002
3\uff0e\u8ba1\u7b97\u516c\u5f0f\uff1a
\uff08\u5f0f\u4e2dc\u4e3a\u6c34\u7684\u6bd4\u70ed\uff0c\u03c1\u4e3a\u6c34\u7684\u5bc6\u5ea6\uff0cU\u4e3a\u7535\u70ed\u58f6\u7684\u989d\u5b9a\u7535\u538b\uff09\u3002
\u4e94\uff0e
\u6c34\u5438\u6536\u7684\u70ed\u91cfQ2=cm\u25b3t=1.0\u00d7500\u00d74.2\u00d7103\u00d7(100-20)\u7126= =168000\uff08\u7126\uff09 (3)
\u8bc4\u5206\u6807\u51c6\uff1a\u5168\u98988\u5206\u3002(1)\u5f0f\u3001(2)\u5f0f\u54042\u5206\u3002(3)\u5f0f3\u5206\u3002(4)\u5f0f1\u5206\u3002
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W=13\u8f6c\u00f7\uff081950\u8f6c\uff0f\u5343\u74e6\u65f6\uff09=13\u00f7\u30141950/(1000\u00d73600)\u3015=24000(\u7126) (1)
\u5176\u5b83\u7528\u7535\u5668\u603b\u529f\u7387P\u5176=n\u706fP\u706f+P\u89c6+P\u54cd=4\u00d760+90+105=435\uff08\u74e6\uff09
=0.435\uff08\u5343\u74e6\uff09 (3)
\u4e00\u6708\u6d88\u8017\u7684\u7535\u80fd
W\u603b=P\u5176t\u5176+P\u51b0t\u51b0
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\u4e03\uff0e\u53c2\u8003\u89e3\u7b54\uff1a\u7535\u70ed\u6c34\u5668\u7684\u7535\u963b
R=32.2\u6b27 121\u79d2\u6240\u505a\u7684\u7535\u529f
W=1.5\u00d7105\u7126 \u5b9e\u9645\u529f\u7387
P\u2019=1240\u74e6
U\u2019=200\u4f0f (7) \u8bc4\u5206\u6807\u51c6 \u5168\u98987\u5206\u3002(1)\uff5e(7)\u5f0f\u6bcf\u5f0f1\u5206\u3002
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1 \u95ed\u5408\u5f00\u5173S\u3001\u8c03\u8282\u6ed1\u52a8\u53d8\u963b\u5668P \u4f7f\u7535\u538b\u8868\u7684\u793a\u6570\u6d88\u8017\u4e3a8V\uff0c\u7535\u6d41\u8868\u7684\u53f2\u4e66\u662f\u591a\u5c11\uff1f\u6b64\u65f6\u6ed1\u52a8\u53d8\u963b\u5668\u6d88\u8017\u7684\u7535\u529f\u7387\u591a\u5927
2 \u82e5\u5728R1\u4e24\u7aef\u5e76\u8054\u4e00\u4e2a\u5b9a\u503c\u7535\u963bR3\u3002\u5728\u8c03\u8282\u6ed1\u52a8\u53d8\u963b\u5668P \u8ba9\u7535\u538b\u8868\u793a\u6570\u4e3a4V\u662f\uff0c\u7535\u6d41\u8868\u7684\u793a\u6570\u4e3a0.75A\uff0c\u6cfd\u3001R3\u7684\u963b\u503c\u4e3a\u591a\u5927 1.I=U/R1=8V/16\u6b27=0.5A
R2=U2/I=4V/0.5A=8\u6b27 P=I*IR=0.5A*0.5A*8\u6b27=2W
2,R1\u7684\u7535\u6d41=4/16=0.25A,R2\u7684\u7535\u6d41=0.75-0.25=0.5A,R3=4/0.5=8\u6b271.\u6709\u4e24\u4e2a\u706f\u6ce1L1\u548cL2 \uff0c\u5206\u522b\u6807\u6709\u201c10V 10W\u201d\u548c\u201c9V 5.4W\u201d\u7684\u5b57\u6837\uff0c\u5f53\u4ed6\u4eec\u4e32\u8054\u8d77\u6765\u4f7f\u7528\uff0c\u7535\u8def\u4e24\u7aef\u5141\u8bb8\u52a0\u7684\u6700\u5927\u7535\u538b\u662f\u591a\u5c11\uff1f\u5f53\u4ed6\u4eec\u5e76\u8054\u4f7f\u7528\u65f6\uff0c\u5e72\u8def\u5141\u8bb8\u901a\u8fc7\u7684\u6700\u5927\u7535\u6d41\u662f\u591a\u5c11\uff1f\u7535\u8def\u6d88\u8017\u7684\u6700\u5927\u529f\u7387\u662f\u591a\u5c11\uff1f
1\u3002
\u77e5\u9053\u989d\u5b9a\u7535\u538b\u8ddf\u989d\u5b9a\u529f\u7387\uff0c\u6839\u636e P=UI \u5f97\u51fa\u989d\u5b9a\u7535\u6d41I1=1\u5b89 I2=0.6\u5b89
U/I=R \u5f97\u51faR1=10\u6b27 R2=15\u6b27
\u4e32\u8054\u7535\u963b\u76f8\u52a0 R\u603b=R1+R2=25\u6b27
\u4e32\u8054\u7684\u65f6\u5019\u7535\u6d41\u6700\u9ad8\u53ea\u80fd\u8fbe\u5230 0.6 \uff0c\u8981\u4e0dL2\u5c31\u7206\u4e86
\u6240\u4ee5\u6700\u5927\u7535\u538b\u4e3a U=R\u603b X 0.6 = 25 X 0.6= 15\u4f0f
\u5e76\u8054\u7684\u8bdd\uff0c\u7535\u538b\u53ea\u80fd\u52a09\u4f0f\uff0c\u8981\u4e0dL2\u8fd8\u662f\u7206\u4e86\uff0c
\u6839\u636eI=U/R \u5f97\u5230\u652f\u8def\u7684I1=9/R1=0.9\u5b89 I2=9/R2=0.6\u5b89
\u5e72\u8def\u7535\u6d41 I\u603b=I1+I2=1.5\u5b89
\u529f\u7387P=UI\u603b=9 X 1.5=13.5
1.\u67d0\u5b9a\u503c\u7535\u963b\u7684\u7ec4\u7ec7\u662f20\u03a9\uff0c\u628a\u5b83\u63a5\u5728\u7535\u8def\u4e2d\uff0c\u5982\u679c\u5b83\u4e24\u7aef\u7684\u7535\u538b\u662f12V\uff0c\u6c42\uff1a
\uff081\uff09\u901a\u8fc7\u5b83\u7684\u5730\u90a3\u516d\u662f\u591a\u5927\uff1f
\uff082\uff09\u7535\u963b\u7684\u7535\u529f\u7387\u662f\u591a\u5927\uff1f
2.\u73b0\u6709\u89c4\u683c\u4e3a\u201c8V 4W\u201d\u7684\u5c0f\u706f\u6ce1\u4e00\u76f4\uff0c\u6c42\uff1a
\uff081\uff09\u5c0f\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c\u65f6\u7684\u7535\u6d41
\uff082\uff09\u706f\u4e1d\u7684\u7535\u963b
\uff083\uff09\u5c0f\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c5min\u6d88\u8017\u7684\u7535\u80fd
3.\u5c06\u4e00\u6807\u6709\u201c220V 880W\u201d\u7684\u7535\u996d\u9505\u4ecb\u5165220V\u7684\u7535\u8def\u4e2d\uff0c\u6c42\uff1a
\uff081\uff09\u901a\u8fc7\u7535\u996d\u9505\u7684\u7535\u6d41\u662f\u591a\u5c11\uff1f
\uff082\uff09\u4f7f\u7528\u6b64\u7535\u996d\u95052min\uff0c\u4e0b\u597d\u7684\u7535\u80fd\u662f\u591a\u5c11\uff1f 1.(1)I=U/R=12V/20\u03a9=0.6A
(2)P=UI=12Vx0.6A=7.2W
2(1)I=P/U=4W/8V=0.5A
(2)R=u^2/P=(8V)^2/4W=16\u03a9
(3)W=Pt=4Wx5x60s=1200J
3(1)I=P/U=880W/220V=4A
(2)W=Pt=880Wx2x60s=105600J1.\u67d0\u5b9a\u503c\u7535\u963b\u7684\u963b\u503c\u662f10\u03a9\uff0c\u628a\u5b83\u63a5\u5728\u7535\u8def\u4e0a\uff0c\u5982\u679c\u5b83\u4e24\u7aef\u7684\u7535\u538b\u662f6V\uff0c\u6c42\uff1a
\uff081\uff09\u901a\u8fc7\u5b83\u7684\u7535\u6d41\u662f\u591a\u5927\uff1f
\uff082\uff09\u7535\u963b\u7684\u7535\u529f\u7387\u662f\u591a\u5927\uff1f
2.\u73b0\u6709\u89c4\u683c\u4e3a\u201c6V 3w\u201d\u7684\u5c0f\u706f\u6ce1\u4e00\u53ea\uff0c\u6c42\uff1a
\uff081\uff09\u5c0f\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c\u65f6\u7684\u7535\u6d41
\uff082\uff09\u706f\u4e1d\u7684\u7535\u963b
\uff083\uff09\u5c0f\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c10min\u6d88\u8017\u7684\u7535\u80fd
3.\u8bb2\u4e00\u6807\u6709\u201c220V 1100W\u201d\u7684\u7535\u5439\u98ce\u63a5\u5165220V\u7684\u7535\u8def\u4e2d\uff0c\u6c42\uff1a
\uff081\uff09\u901a\u8fc7\u7535\u5439\u98ce\u7684\u7535\u6d41\u662f\u591a\u5c11\uff1f
\uff082\uff09\u4f7f\u7528\u6b21\u7535\u5439\u98ce2min\uff0c\u6d88\u8017\u7535\u80fd\u662f\u591a\u5c11\uff1f
1(1)I=U/R=6/10=0.6A
(2)P=I^2*R=3.6w
2(1)I=P/U=0.5A
(2)R=U/I=12
(3)W=Pt=3*10*60=1800
3(1)I=P/U=5A
(2)W=Pt=1100*2*60=13200 1.\u5bb6\u5ead\u7535\u8def\u4e2d\u7684\u7535\u51b0\u7bb1\u6b63\u5e38\u5de5\u4f5c\u65f6\u7684\u7535\u6d41\u662f0.7A\uff0c\u5219\u5b83\u7684\u7535\u529f\u7387\u662f\u591a\u5c11\uff1f\u82e5\u8ba9\u4ed6\u5355\u72ec\u5de5\u4f5c1h,\u5b83\u6d88\u8017\u4e86\u591a\u5c11\u7535\u80fd?
2.\u5bb6\u5ead\u7535\u8def\u4e2d200w\u7684\u706f\u6ce1\u57283min\u5185,\u6d88\u8017\u7684\u7535\u80fd\u662f\u591a\u5c11kw.h?\u6d41\u8fc7\u706f\u6ce1\u7684\u7535\u6d41\u662f\u591a\u5c11?\u5c0f\u706f\u6ce1\u7684\u7535\u963b\u662f\u591a\u5c11?
3.\u706f\u6ce1\u4e0a\u6807\u6709PZ 220 40\u7684\u5b57\u6837.\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c\u65f6,\u6d41\u8fc7\u5b83\u7684\u7535\u6d41\u7ea6\u662f\u591a\u5c11?\u5c0f\u706f\u6ce1\u7684\u7535\u963b\u662f\u591a\u5927?\u82e5\u5c0f\u706f\u6ce1\u6b63\u5e38\u5de5\u4f5c1min,\u6d88\u8017\u4e86\u591a\u5c11\u7535\u80fd?
4.\u674e\u521a\u540c\u5b66\u5bb6\u7684\u7535\u80fd\u8868\u4e0a\u6807\u6709600r/kw.h\u7684\u5b57\u6837,\u636e\u6b64,\u4ed6\u7528\u8868\u8bb0\u5f55\u4e8610min\u5185\u7535\u80fd\u8868\u4e0a\u7684\u8f6c\u76d8\u8f6c\u8fc7120r.\u90a3\u4e48\u7528\u7535\u5668\u6240\u6d88\u8017\u7684\u7535\u80fd\u662f\u591a\u5c11?\u6b64\u7528\u7535\u5668\u7684\u7535\u529f\u7387\u662f\u591a\u5c11?
1\u3001P=IU=0.7A*220v=154w=0.154KW W=pt=0.154Kw*1h=0.154kwh
2\u3001200w=0.2kw 3min=1/20h w=Pt=0.2kw*1/20h=0.01kwh I=P/U=200w/220v=10/11A
R=U/I=220v/(10/11A)=242\u6b27
3\u3001I=P\u989d/U\u989d=40w/220v=2/11A R=U\u989d^2/P\u989d=(220v)^2/40w=4840\u6b27 W=P\u989dt=40w*60s=2400J
4\u300110min(1/6h)\u5185\u7535\u80fd\u8868\u6240\u6d88\u8017\u7684\u7535\u80fdW=120r/(600r/kw.h)=0.2kw.h
P=W/t=0.2kw.h/(1/6h)=1.2kw
1 \u3001 \u706f\u6ce1L 1 \u548cL 2 \u4e32\u8054\u5728\u7535\u8def\u4e2d\uff0cL 1 \u7684\u7535\u963b\u6bd4L 2 \u5927\uff0c\u5728\u76f8\u7b49\u7684\u65f6\u95f4\u91cc\uff0c\u7535\u6d41\u505a\u7684\uff08 \uff09 A\u3001L 1 \u706f\u505a\u7684\u529f\u591a B\u3001L 2 \u706f\u505a\u7684\u529f\u591a C\u3001\u4e24\u706f\u4e00\u6837\u591a D\u3001\u65e0\u6cd5\u786e\u5b9a \u89e3\uff1a \u4e24\u706f\u4e32\u8054\uff0c\u901a\u8fc7\u4e24\u706f\u7684\u7535\u6d41\u76f8\u540c\uff0c\u65f6\u95f4\u4e5f\u76f8\u7b49\uff0c\u7531W=UIt\u548cU=IR\u5f97W=I 2 Rt\u53ef\u77e5\u7535\u529f\u4e0e\u7535\u963b\u6210\u6b63\u6bd4\uff0c\u800cL 1 \u7684\u7535\u963b\u5927\u4e8eL 2 \u7684\u7535\u963b\uff0c\u6240\u4ee5\u7535\u6d41\u901a\u8fc7L 1 \u505a\u529f\u591a\uff0c\u672c\u9898\u5e94\u9009A\u3002 2 \u3001 \u628a\u201c12V\u30018W\u201d\u7684\u7535\u706f\u63a5\u52309V\u7684\u7535\u8def\u4e2d\uff0c\u5b83\u7684\u5b9e\u9645\u529f\u7387\u4e3a\u591a\u5927\uff1f \u89e3\uff1a 3 \u3001 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A\u3001\u5c06\u7535\u7089\u4e1d\u622a\u6389\u4e00\u534a\u540e\u63a5\u5165\u7535\u8def B\u3001\u5c06\u7535\u7089\u4e1d\u622a\u6389\u56db\u5206\u4e4b\u4e09\u63a5\u5165\u7535\u8def C\u3001\u5c06\u7535\u7089\u4e1d\u5206\u6210\u7b49\u957f\u7684\u4e24\u6bb5\uff0c\u7136\u540e\u5e76\u8054\u63a5\u5165\u7535\u8def D\u3001\u5c06\u7535\u7089\u4e1d\u622a\u6389\u56db\u5206\u4e4b\u4e00\u540e\u63a5\u5165\u7535\u8def \u89e3\uff1a \u6839\u636e\u9898\u610f\u8981\u6c42 \u65f6\uff0cP \u5b9e =P \u989d \u3002\u7531 \u53ef\u77e5\u8981\u6ee1\u8db3\u9898\u8bbe\u6761\u4ef6\uff0c\u53ea\u6709\u6539\u53d8\u7535\u963bR\uff0c\u624d\u80fd\u6ee1\u8db3\u8981\u6c42\u3002\u7531 \u82e5\uff0cP \u5b9e \u624d\u80fd\u7b49\u4e8eP \u989d \u3002\u6240\u4ee5 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如图,R1=200欧，R2=50欧，电源电压不变。（1）、当开关S1闭合，S2断开时，电流表示数为0.1安电压表示数为12伏，求滑动变阻器连入电路的电阻值和电源电压。
（2）、当开关S1、S2都闭合时，移动滑动变阻器滑片P使电压表示数为零时，求电流表示数
.
（1）由图可知电压表所测电压是滑动变阻器的，再根据串联电路中电流处处相等得
R阻=U/I=12v/0.1A=120欧
U=（R1+R阻）/I=(200欧+120欧）*0.1A=32v
(2)由题可知，滑动变阻器的阻值为0，根据并联电路的特点R总=R1R2/R1+R2得
R总=200欧*50欧/200欧+50欧=40欧
I=U/R=32v/40欧=0.8A
1.已知R1和R2的电阻之比为3:5,而加在R1、R2两端的电压之比为2:1,通过他们的电流之比为_________________
2.两个完全相同的电阻,它们串联的总电阻是并联总电阻的_________倍?
3.定值电阻R1、R2和滑动变阻器R串联，R1、R2两端分压分别为U1和U2，当滑动变阻器的阻值变大时U1变小，则U1：U2____________(变大、变小、不变)
4.有三只电阻，R1＝10欧姆，R2＝5欧姆，R3＝1欧姆怎样把它们连接起来，它们的值最小？
A 全部串联 B全部并联 C R1与R2并联再与R3串联 D R1与R3并联再与R2串联
5.有一个功率为100KW的电力用户，所需电压为400V，如果不用变压器而直接输电，输电线路电阻消耗的功率为62.5KW，要让损失减到100W，输电电压应提高到多少伏？
6.将三根电阻丝并联在某一电路间，若R1:R2:R3=1:2:3,则三根电阻丝消耗的功率比为多少？
7.1A约是下列哪个的的电流值?
A.收音机中的电流 B.手电筒中的电流
C.家用电冰箱的电流 D.雷击电流
8.R1 R2以某种方法连接
R1的电功率12W R2的电功率为P2
若将它们以另一种方法连接起来接入原电路中
此时 R1的电功率108W R2的电功率P2'
求P2 P2'的值
9.如果说:一个物体带正电荷,那么,是说这个物体仅仅带有正电荷,还是它的正电荷的数量比负电荷的数量多呢?
10.电源电压为15V保持不变，灯泡上只标有“6V'字样，变阻器R1
的最大阻值为36欧姆，闭合开关S后，滑片置于中点时，小灯泡恰好正常发光，根据以上数据，计算出个电学物理量。
注：电源，开关S，滑动变阻器R1，灯泡L都是串联的。
“PZ220—25”和“PZ220—40”的甲、乙两灯泡串联后接在220伏的电路中，则甲灯比乙灯 。（填“亮”或“暗”），若通过两灯的总电量为20库，则电流通过两灯共做功是 多少焦。
答案：⒈ ∵R1：R2=3:5 U1:U2=2:1 I=U/R ∴I1:I2=6:5
⒉ 设R=1Ω R串=2Ω R并= R×R/(R+R)=0.5Ω
∴R串：R并=2：1
⒊ U1:U2 U1↓ U1:U2↓
⒋ 电路并联时电阻最小 总电功率最大
⒌我没看懂
⒍ ∵并联 ∴电压相等 P=U的平方/R
∵R1:R2:R3=1:2:3 ∵电压相等
∴P1:P2:P3=3:2:1
⒎ 这个不太清楚 ABD都不对 应该是C
⒏ ∵电功率变大了 ∴第一次串联第二次并联
P=U方/R R1电阻不变 则U串R1:U并R1=1:3
（因为是U平方 所以 1：9要开方）
所以R2电阻是R1两倍 ∵P=I方R 串联I等
R1:R2=1:2 P1:P2=1:2 ∴P2=24W
串联U等 P=U方/R R1:R2=1:2 P1:P2=2:1
∵P1=108W ∴P2'=54W
⒐ 一般都是带正电荷多 好像还没有不带负电荷的
⒑ 滑片在中点时电阻一半18Ω ∵正常发光时电阻18Ω
灯泡额定电压是6V 因为都是串连 所以
R小灯泡：R滑=6：（15-6）=2：3
R滑=18Ω R小灯泡=18乘3分之2=12Ω 正常发光时
R縂=30Ω U縂=15V I縂=U/R=15/30=0.5A
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