The voltage drop across the 15, Omega resista | vedclass

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Question

(C) First, simplify the circuit. The parallel combination of two $4\, \Omega$ resistors is $2\, \Omega$. The parallel combination of two $8\, \Omega$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) First, simplify the circuit. The parallel combination of two $4\, \Omega$ resistors is $2\, \Omega$. The parallel combination of two $8\, \Omega$ resistors is $4\, \Omega$. The parallel combination of two $12\, \Omega$ resistors is $6\, \Omega$.The upper branch has resistances in series: $2\, \Omega$ (from $4\, \Omega || 4\, \Omega$) + $2\, \Omega$ + $R_{p1}$ (where $R_{p1} = 15\, \Omega || 10\, \Omega = 6\, \Omega$). Total upper branch resistance $R_1 = 2 + 2 + 6 = 10\, \Omega$.The lower branch has resistances in series: $4\, \Omega$ (from $8\, \Omega || 8\, \Omega$) + $6\, \Omega$ (from $12\, \Omega || 12\, \Omega$) = $10\, \Omega$.The two branches are in parallel, so their equivalent resistance is $R_{eq_branches} = (10\, \Omega || 10\, \Omega) = 5\, \Omega$.Total circuit resistance $R_{total} = 5\, \Omega + 1\, \Omega$ (internal) $= 6\, \Omega$.Total current $I = V / R_{total} = 12\, V / 6\, \Omega = 2\, A$.This current splits equally into the two $10\, \Omega$ branches, so $1\, A$ flows through the upper branch.The voltage across the $15\, \Omega || 10\, \Omega$ parallel combination is $V_{AB} = I_{upper} 	imes R_{p1} = 1\, A 	imes 6\, \Omega = 6\, V$.

List-$I$	List-$II$
$(I)$ Two wires of same resistance are connected in series and same current is passed through them	$(A)$ $1:2$
$(II)$ Two wires of resistance $R$ and $2R$ $\Omega$ are connected in series and same $P.D.$ is applied across them	$(B)$ $4:1$
$(III)$ Two wires of same resistance are connected in parallel and same current is flowing through them	$(C)$ $1:1$
$(IV)$ Two wires of resistances in the ratio $1:2$ are connected in parallel and same $P.D.$ is applied across them	$(D)$ $2:1$

The voltage drop across the $15\, \Omega$ resistance in the given figure is $.....V.$

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