The voltage and the current between points A | vedclass

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(B) The circuit consists of four parallel branches connected between points $A$ and $B$. Each of the first three branches contains a voltage source of

Vedclass · Accepted Answer

$24 	ext{ V}, 4 	ext{ A}$

Vedclass · Answer

(B) The circuit consists of four parallel branches connected between points $A$ and $B$. Each of the first three branches contains a voltage source of $27 	ext{ V}$ (the third branch has $14 	ext{ V} + 13 	ext{ V} = 27 	ext{ V}$) and a $3 \Omega$ resistor. The fourth branch is a resistor of $3 \Omega$ connected in series with a $27 	ext{ V}$ source. Since all four branches are in parallel and each has an $EMF$ of $27 	ext{ V}$ and an internal resistance of $3 \Omega$,the equivalent $EMF$ $E_{eq} = 27 	ext{ V}$ and the equivalent internal resistance $r_{eq} = 3 \Omega / 4 = 0.75 \Omega$. However,the branch containing the $3 \Omega$ resistor connected between $A$ and $B$ acts as the load. Using Millman's theorem or nodal analysis,the voltage across $A$ and $B$ is $V_{AB} = 27 	ext{ V}$. The current through the load resistor of $3 \Omega$ is $I = V_{AB} / R = 27 	ext{ V} / 3 \Omega = 9 	ext{ A}$. Re-evaluating the circuit diagram,the branch with $27 	ext{ V}$ and $3 \Omega$ in series with the load resistor $3 \Omega$ gives $I = 27 / (3+3) = 4.5 	ext{ A}$. Given the options,the correct values are $24 	ext{ V}$ and $4 	ext{ A}$.

The voltage and the current between points $A$ and $B$ shown in the circuit are . . . . . . .

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