A current of 2 , A flows in an electric circu | vedclass

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(C) The circuit consists of two parallel branches connected between points $P$ and $Q$. Branch $PRQ$ has a total resistance $R_1 = 3 \, \Omega + 7 \,

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$+4$

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(C) The circuit consists of two parallel branches connected between points $P$ and $Q$. Branch $PRQ$ has a total resistance $R_1 = 3 \, \Omega + 7 \, \Omega = 10 \, \Omega$. Branch $PSQ$ has a total resistance $R_2 = 7 \, \Omega + 3 \, \Omega = 10 \, \Omega$. Since the resistances of both branches are equal, the total current of $2 \, A$ splits equally into $1 \, A$ through each branch. For branch $PRQ$, the current $I_1 = 1 \, A$. The potential drop across the $3 \, \Omega$ resistor is $V_P - V_R = I_1 	imes 3 \, \Omega = 1 \, A 	imes 3 \, \Omega = 3 \, V$. For branch $PSQ$, the current $I_2 = 1 \, A$. The potential drop across the $7 \, \Omega$ resistor is $V_P - V_S = I_2 	imes 7 \, \Omega = 1 \, A 	imes 7 \, \Omega = 7 \, V$. We want to find $V_R - V_S$. From the above equations: $V_R = V_P - 3$ $V_S = V_P - 7$ Subtracting the two equations: $V_R - V_S = (V_P - 3) - (V_P - 7) = -3 + 7 = +4 \, V$.

$A$ current of $2 \, A$ flows in an electric circuit as shown in the figure. The potential difference $(V_R - V_S)$, in volts ($V_R$ and $V_S$ are potentials at $R$ and $S$ respectively) is:

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