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0.5 volt

3.2 volt

1.5 volt

1.0 volt

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Step-by-Step Solution Step 1: Identify the parallel combination The 4 Ω and 3 Ω resistors are connected in parallel. This means they share the same potential difference (P.D.). Step 2: Determine the current distribution in the parallel branch Let the current through the 4 Ω resistor be 1 A (as given or assumed in the solution), and let the current through the 3 Ω resistor be $i_1$. Since the potential difference across these two resistors is the same, we can equate: $4 \times 1 \, \Omega = 3 \times i_1 \,$ Solving for $i_1$: $i_1 = \frac{4}{3} \, \text{A}$ Step 3: Compute the equivalent resistance of the parallel branch The combined resistance of two parallel resistors $R_1$ and $R_2$ is given by $R_{\text{parallel}} = \frac{R_1\,R_2}{R_1 + R_2}$. Here, $R_1 = 4\,\Omega$ and $R_2 = 3\,\Omega$, so: $ R_{\text{parallel}} = \frac{4 \times 3}{4 + 3} = \frac{12}{7} \,\Omega $ Step 4: Calculate the total current through the upper branch (MQP) In that branch, the total current is the sum of currents through the 4 Ω resistor (1 A) and the 3 Ω resistor ($\frac{4}{3}$ A). Thus: $ I_{\text{MQP}} = 1 + \frac{4}{3} = \frac{7}{3} \,\text{A} $ Step 5: Determine the potential difference across the parallel branch The potential difference across the combined parallel resistance $\frac{12}{7}\,\Omega$ with the current $\frac{7}{3}\,\text{A}$ is: $ V_{\text{parallel}} = R_{\text{parallel}} \times I_{\text{MQP}} = \left(\frac{12}{7}\right) \times \left(\frac{7}{3}\right) = 4 \,\text{V} $ Step 6: Find the current in the MNP branch According to the solution, the current in branch MNP (through a 1 Ω resistor and possibly other components with total resistance 1.25 Ω) is given by: $ I_{\text{MNP}} = \frac{4}{1.25} = \frac{4 \times 4}{5} = \frac{16}{5} \,\text{A} $ Step 7: Calculate the potential difference across the 1 Ω resistor Finally, we find the voltage drop across the 1 Ω resistor in the MNP path by using Ohm’s law $V = IR$: $ V_{1\,\Omega} = I_{\text{MNP}} \times 1\,\Omega = \frac{16}{5} \times 1\,\Omega = \frac{16}{5} \,\text{V} = 3.2 \,\text{V} $ Final Answer: 3.2 V The potential difference across the 1 Ω resistor is therefore 3.2 V.

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Parent Quiz 802591

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