A current of 2,A flows in a system of conduct | vedclass

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(B) The circuit consists of two parallel branches connected between points $D$ and $C$. Branch $DAC$ has a total resistance of $R_1 = 2\,\Omega + 3\,\

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(B) The circuit consists of two parallel branches connected between points $D$ and $C$. Branch $DAC$ has a total resistance of $R_1 = 2\,\Omega + 3\,\Omega = 5\,\Omega$. Branch $DBC$ has a total resistance of $R_2 = 3\,\Omega + 2\,\Omega = 5\,\Omega$. Since the branches are in parallel and have equal resistance,the total current of $2\,A$ splits equally. Thus,current $I_1$ through branch $DAC$ is $1\,A$ and current $I_2$ through branch $DBC$ is $1\,A$. Using Ohm's law,the potential at $A$ relative to $D$ is $V_D - V_A = I_1 	imes 2\,\Omega = 1\,A 	imes 2\,\Omega = 2\,V$. The potential at $B$ relative to $D$ is $V_D - V_B = I_2 	imes 3\,\Omega = 1\,A 	imes 3\,\Omega = 3\,V$. To find $(V_A - V_B)$,we subtract the two equations: $(V_D - V_B) - (V_D - V_A) = 3\,V - 2\,V$ $V_A - V_B = 1\,V$.

$A$ current of $2\,A$ flows in a system of conductors as shown. The potential difference $(V_A - V_B)$ will be ............ $V$.

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