In the given circuit,if the current flowing t | vedclass

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(A) $1$. The $10 \Omega$ and $10 \Omega$ resistors are in parallel,so their equivalent resistance is $R_1 = (10 	imes 10) / (10 + 10) = 5 \Omega$.$2$

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$6$

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(A) $1$. The $10 \Omega$ and $10 \Omega$ resistors are in parallel,so their equivalent resistance is $R_1 = (10 	imes 10) / (10 + 10) = 5 \Omega$.$2$. This $5 \Omega$ is in series with the $3 \Omega$ resistor,so $R_2 = 5 + 3 = 8 \Omega$.$3$. This $8 \Omega$ branch is in parallel with the $8 \Omega$ resistor,so $R_3 = (8 	imes 8) / (8 + 8) = 4 \Omega$.$4$. The $6 \Omega$ and $6 \Omega$ resistors are in parallel,so $R_4 = (6 	imes 6) / (6 + 6) = 3 \Omega$.$5$. The total equivalent resistance of the external circuit is $R_{eq} = R_3 + R_4 = 4 + 3 = 7 \Omega$.$6$. The voltage across the $5 \Omega$ resistor is $V = I 	imes R = 0.5 	ext{ A} 	imes 5 \Omega = 2.5 	ext{ V}$. This is the terminal voltage of the battery.$7$. The total current $I_{total}$ flowing through the battery is $V / R_{eq} = 2.5 / 7 \approx 0.357 	ext{ A}$.$8$. Using $E = V + I_{total} 	imes r = 2.5 + (0.357 	imes 2) = 2.5 + 0.714 = 3.214 	ext{ V}$.$9$. Re-evaluating the circuit diagram: The $5 \Omega$ resistor is in parallel with the rest of the network. The voltage across the $5 \Omega$ resistor is $2.5 	ext{ V}$. The rest of the network also has $2.5 	ext{ V}$ across it. The current through the rest of the network is $I_{net} = 2.5 / 7 \approx 0.357 	ext{ A}$. Total current $I = 0.5 + 0.357 = 0.857 	ext{ A}$. $E = 2.5 + 0.857 	imes 2 = 2.5 + 1.714 = 4.214 	ext{ V}$. The closest option is $4 	ext{ V}$.

In the given circuit,if the current flowing through the $5 \Omega$ resistor is $0.5 \text{ A}$,then the value of $E$ is: (in $\text{ V}$)

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