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(A) $1$. Analyze the circuit: The $10\,\Omega$ and $15\,\Omega$ resistors are in parallel,and the voltage across them is $10\,V$. $2$. Current in $10\

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$20\,V, 2\,A$

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(A) $1$. Analyze the circuit: The $10\,\Omega$ and $15\,\Omega$ resistors are in parallel,and the voltage across them is $10\,V$. $2$. Current in $10\,\Omega$ resistor: $I_{10} = \frac{V}{R} = \frac{10\,V}{10\,\Omega} = 1\,A$. $3$. Current in $15\,\Omega$ resistor: $I_{15} = \frac{10\,V}{15\,\Omega} = \frac{2}{3}\,A$. $4$. Total current through the middle branch: $I_{total} = 1 + \frac{2}{3} = \frac{5}{3}\,A$. $5$. Equivalent resistance of the network: The circuit simplifies to a series combination of $4\,\Omega$,the parallel block,and $2\,\Omega$. The parallel block consists of $(10\,\Omega || 15\,\Omega) = 6\,\Omega$ in series with $3\,\Omega$ and $3\,\Omega$ (which are in parallel with $6\,\Omega$). Calculating the total equivalent resistance $R_{eq} = 4 + 6 + 2 = 12\,\Omega$. $6$. Potential difference $V_{AB} = I_{total} 	imes R_{eq} = \frac{5}{3}\,A 	imes 12\,\Omega = 20\,V$.

Calculate the potential difference between points $A$ and $B$ and the current flowing in the $10\,\Omega$ resistor in the network shown below.

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