1.

A current of 3 A flows through the 2Omegaresistor shown in the circuit. The power dissipated in the 5 Omegaresistor is

Answer»

5 WATT.
4 watt.
2 watt.
1 watt

Solution :In the circuit arrangement RESISTANCES `R_1 = 2Omega, R_2 = 4 Omega` and series COMBINATION of `1 Omega` and `5Omega`resistances i.e., `R_3=1+5=6Omega`are joined in parallel. If CURRENT through 6 `Omega` resistance combination be `I_3` then `I_1 R_1 = I_3 R_3`
` therefore I_3 = (I_1R_1)/(R_3) = (3 xx 2)/(6) = 1A`
Power dissipated in 5 `Omega`RESISTOR` P= I_3^2R = (1)^2 xx 5 = 5`watt


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