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An electric heater of resistance 10Ω and resistance of the wine 8Ω are connected in series with a 6 volt battery find. 1. Current through circuits 2. Potential across the heater3. Potential across the wire |
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Answer» 10Ω resistance and 8Ω resistance are connected in series. R = R1 + R2 R = 10 + 8 R = 18Ω (i) Current through circuit V = IR I = \(\frac{V}R\) I = 6/18 I = \(\frac13\) ampeare (ii) Potential across the heater V = IR V = 1/3 x 10 V = 10/3 v (iii) Potential across the wire V = IR V = 1/3 x 8 V = 8/3 v |
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