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NCERT

Class 10

Science

Solutions

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ωresistor?

The resistance is connected in series

R_{S} = 0.2Ω + 0.3Ω + 0.4Ω + 0.5Ω + 12Ω

R_{S} = 13.4Ω

Current = I = V/R_{S}

I = 9/13.4

I = 0.67A

So, the current flowing through 12 W resistor = 0.67 A.

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