A current of 3 amp. flows through the 2 Ω resistor shown in the circuit. The power dissipated in the 5 Ω resistor is
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Current flows through the 9 Ω resistor is I^2 = 36/9 = 4. As, P1 = I1^2R. I1 = 2A.
The resistors 9 Ω and 6 Ω are in parallel connection.
As a result the potential difference across the resistors is same. 9I1 = 6I2 or, I2 = 9*2/ 6 = 3A.
The current drawn from the battery would be I = I1 +I2 = (2 + 3 ) = 5A.
Hence the potential difference is 5 * 2 = 10 V.
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