The reading of the ideal voltmeter in the adj | vedclass

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(B) The circuit consists of two batteries of $10\,V$ and $4\,V$ connected in opposition,and two resistors of $10\,\Omega$ and $20\,\Omega$ in series.T

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The circuit consists of two batteries of $10\,V$ and $4\,V$ connected in opposition,and two resistors of $10\,\Omega$ and $20\,\Omega$ in series.Total electromotive force $(EMF)$ in the circuit is $E_{net} = 10\,V - 4\,V = 6\,V$.Total resistance in the circuit is $R_{total} = 10\,\Omega + 20\,\Omega = 30\,\Omega$.The current flowing in the circuit is $i = \frac{E_{net}}{R_{total}} = \frac{6\,V}{30\,\Omega} = 0.2\,A = \frac{1}{5}\,A$.The voltmeter is connected between points $A$ and $N$. We can calculate the potential difference $V_{AN}$ by traversing the path $A-B-N$.Moving from $A$ to $B$,we pass through the $10\,V$ battery (from positive to negative terminal),so the potential drops by $10\,V$.Moving from $B$ to $N$,we pass through the $10\,\Omega$ resistor in the direction of current,so the potential drops by $iR = (0.2\,A) 	imes (10\,\Omega) = 2\,V$.Thus,$V_A - V_N = 10\,V - 2\,V = 8\,V$.Alternatively,traversing path $A-C-N$: $V_A - V_C = i 	imes (20\,\Omega) = 0.2 	imes 20 = 4\,V$. Moving from $C$ to $N$,we pass through the $4\,V$ battery (from negative to positive terminal),so the potential increases by $4\,V$. Thus,$V_A - V_N = 4\,V + 4\,V = 8\,V$.

The reading of the ideal voltmeter in the adjoining diagram will be ................. $V$.

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