Answer :

(a) Here V = 6 volt, R_{1} = 1 ohm, R_{2} = 2 ohm

The connection is in series. So Equivalent resistance = R_{1}+R_{2} = 1 + 2 = 3 ohm

We know that

Total current I = V/R = 6/3 = 2A

Current through R_{2} = I_{2} = I = 2A

Voltage across R_{2} = V_{2} = I_{2}R_{2} = 2×2 = 4 ohm

Power used in R_{2} = I_{2}V_{2} = 2×4 = 8W

(b) Here, V = 4 volt,R_{1} = 12 ohm, R_{2} = 2 ohm

We know that, as the connection is in parallel so voltage across R_{2} = V_{2} = V = 4V

Current across R_{2} = I_{2} = V_{2}/R_{2} = 4/2 = 2A

As power = IV

Power used in R_{2} = I_{2}V_{2} = 2×4 = 8W

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