A current of 15,mA flows in the circuit as sh | vedclass

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(D) The circuit consists of a $5\,k\Omega$ resistor in series with a parallel combination of a $10\,k\Omega$ resistor and a $5\,k\Omega$ resistor,foll

Vedclass · Accepted Answer

$275$

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(D) The circuit consists of a $5\,k\Omega$ resistor in series with a parallel combination of a $10\,k\Omega$ resistor and a $5\,k\Omega$ resistor,followed by another $10\,k\Omega$ resistor in series.First,calculate the equivalent resistance of the parallel part $(R_p)$:$R_p = \frac{10\,k\Omega 	imes 5\,k\Omega}{10\,k\Omega + 5\,k\Omega} = \frac{50}{15}\,k\Omega = \frac{10}{3}\,k\Omega$.The total current $I = 15\,mA$ flows through the $5\,k\Omega$ resistor,then splits into the parallel branches,and finally flows through the $10\,k\Omega$ resistor.The potential difference $V_{AB}$ is the sum of the potential drops across each component in the path from $A$ to $B$:$V_{AB} = I 	imes R_{5k} + I 	imes R_p + I 	imes R_{10k}$$V_{AB} = (15\,mA 	imes 5\,k\Omega) + (15\,mA 	imes \frac{10}{3}\,k\Omega) + (15\,mA 	imes 10\,k\Omega)$$V_{AB} = 75\,V + 50\,V + 150\,V = 275\,V$.

$A$ current of $15\,mA$ flows in the circuit as shown in the figure. The value of the potential difference between the points $A$ and $B$ will be $...\,V$.

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