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An electric lamp of resistance `2 Omega` and a conductor of resistance `4 Omega`are connected to a `6 V` Battery as shown in the circuit. Calculate: (a) the totak resistance of the circuit, (b) the current through the circuit, (c) the potential difference across the (i) electric lamp and (ii) conductor, and (d) power of the lamp. |
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Answer» `R_(1)=20 Omega, R_(2) = 4 Omega, V=6V` (a) Total resistance, `R= R_(1) + R_(2) = 20 +4 = 24 Omega` (b) Current, `I=V/R = 6/24=1/4=0.25A` (c) Potential difference across (i) Electric lamp, `V_(1)=IR_(1)=0.25xx 20 =5 "volts"` (ii) Conductor, `V_(2)=IR_(2)=0.25xx 4 =1 "volt"` (d) Power of the lamp, `P=VI=6xx0.25=1.5 W … where[V=V_(1)+V_(2)]` |
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