In the circuit shown,the point 'B' is earthed | vedclass

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(B) The total resistance of the circuit is $R_{eq} = 5\,\Omega + 7\,\Omega + 10\,\Omega + 3\,\Omega = 25\,\Omega$.The current flowing through the circ

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$24$

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(B) The total resistance of the circuit is $R_{eq} = 5\,\Omega + 7\,\Omega + 10\,\Omega + 3\,\Omega = 25\,\Omega$.The current flowing through the circuit is $i = \frac{V}{R_{eq}} = \frac{50\,V}{25\,\Omega} = 2\,A$.Since point '$B$' is earthed,its potential is $V_B = 0\,V$.Moving from '$B$' to '$A$' against the direction of current,the potential increases across the resistors. The resistance between '$A$' and '$B$' is $R_{AB} = 5\,\Omega + 7\,\Omega = 12\,\Omega$.The potential difference is $V_A - V_B = i 	imes R_{AB} = 2\,A 	imes 12\,\Omega = 24\,V$.Since $V_B = 0\,V$,we have $V_A = 24\,V$.

Order	$V \ (V)$	$I \ (A)$
$1$	$1.0$	$0.08$
$2$	$0.5$	$0.18$
$3$	$0.8$	$0.12$

In the circuit shown,the point '$B$' is earthed. The potential at the point '$A$' is ............. $V$.

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