Refer to the figure given below,current betwe | vedclass

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(C) The circuit consists of four parallel branches connected between terminals $A$ and $B$.$1$. The top three branches are identical. Each contains th

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$5$

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(C) The circuit consists of four parallel branches connected between terminals $A$ and $B$.$1$. The top three branches are identical. Each contains three $5	ext{ V}$ batteries in series (total $EMF$ = $5 + 5 + 5 = 15	ext{ V}$) and three $3	ext{ }\Omega$ resistors in series (total resistance = $3 + 3 + 3 = 9	ext{ }\Omega$).$2$. The current in each of these three branches is $I_1 = V/R = 15	ext{ V} / 9	ext{ }\Omega = 5/3	ext{ A}$.$3$. The bottom branch contains only three $3	ext{ }\Omega$ resistors in series (total resistance = $9	ext{ }\Omega$) and no battery. The potential difference across this branch is the same as the others,which is $15	ext{ V}$ (due to the parallel connection).$4$. The current in the bottom branch is $I_2 = V/R = 15	ext{ V} / 9	ext{ }\Omega = 5/3	ext{ A}$.$5$. The total current flowing between terminals $A$ and $B$ is the sum of the currents in all four branches: $I_{total} = I_1 + I_1 + I_1 + I_2 = 4 	imes (5/3) = 20/3 \approx 6.67	ext{ A}$.Wait,re-evaluating the circuit: The batteries are in series with resistors. Each of the top three branches has a net $EMF$ of $15	ext{ V}$ and resistance of $9	ext{ }\Omega$. The bottom branch has $0	ext{ V}$ and $9	ext{ }\Omega$. Since they are in parallel,the potential difference $V_{AB}$ is determined by the combination. Using Millman's theorem: $V_{AB} = \frac{\sum (E/R)}{\sum (1/R)} = \frac{(15/9 + 15/9 + 15/9 + 0/9)}{(1/9 + 1/9 + 1/9 + 1/9)} = \frac{45/9}{4/9} = 45/4 = 11.25	ext{ V}$.The total current $I = V_{AB} / R_{eq}$,where $R_{eq} = 9/4 = 2.25	ext{ }\Omega$. Thus,$I = 11.25 / 2.25 = 5	ext{ A}$.

Refer to the figure given below,current between terminals $A$ and $B$ is . . . . . . $A$.

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