MHT CET · Physics · Electrostatics
Potential difference between the points \(\mathrm{P}\) and \(\mathrm{Q}\) is nearly
- A \(6 \mathrm{~V}\)
- B \(8 \mathrm{~V}\)
- C \(17 \mathrm{~V}\)
- D \(21 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(17 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
As, the resistors across each branch are in series.
\(
\begin{aligned}
& \mathrm{R}_1=6+3=9 \Omega \\
& \mathrm{R}_2=8+4=12 \Omega
\end{aligned}
\)
According to \(\mathrm{KCL}\), the current (i) will get divided inte two parts \(\mathrm{I}_1\) and \(\mathrm{I}_2\)
\(
\mathrm{I}_1=\frac{\mathrm{R}_2}{\left(\mathrm{R}_1+\mathrm{R}_2^{\prime}\right) \mathrm{i}}=\frac{12}{9+12} \times 5=2.86 \mathrm{~A}
\)
Potential difference between \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{V}=\mathrm{I}_1 \mathrm{R}=2,86 \times 6=17.14 \mathrm{~V}\)
\(
\begin{aligned}
& \mathrm{R}_1=6+3=9 \Omega \\
& \mathrm{R}_2=8+4=12 \Omega
\end{aligned}
\)
According to \(\mathrm{KCL}\), the current (i) will get divided inte two parts \(\mathrm{I}_1\) and \(\mathrm{I}_2\)
\(
\mathrm{I}_1=\frac{\mathrm{R}_2}{\left(\mathrm{R}_1+\mathrm{R}_2^{\prime}\right) \mathrm{i}}=\frac{12}{9+12} \times 5=2.86 \mathrm{~A}
\)
Potential difference between \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{V}=\mathrm{I}_1 \mathrm{R}=2,86 \times 6=17.14 \mathrm{~V}\)
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