MHT CET · Physics · Electrostatics
Potential difference between the points \(P\) and Q is nearly
- A \(17 \mathrm{~V}\)
- B \(14 \mathrm{~V}\)
- C \(12 \mathrm{~V}\)
- D \(8 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(17 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let total current be denoted as I. The given circuit is a Wheatstone bridge.
\(\begin{aligned}
& \Rightarrow R_1=6+3=9 \Omega . \\
& \Rightarrow R_2=8+4=12 \Omega
\end{aligned}\)
According to \(\mathrm{KCL}\), the current will get divided into two parts \(I_1\) and \(I_2\)
\(\therefore \quad \mathrm{I}_1=\frac{\mathrm{R}_2 \mathrm{I}}{\left(\mathrm{R}_1+\mathrm{R}_2\right)}\)
Substituting the values,
\(\begin{aligned}
& I_1=\frac{12}{9+12} \times 5 \\
& I_1=2.85 \mathrm{~A}
\end{aligned}\)
Potential difference between \(P\) and \(Q\) is
\(\begin{aligned}
& V=I_1 R \\
& V=2.85 \times 6 \\
& V=17 V
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow R_1=6+3=9 \Omega . \\
& \Rightarrow R_2=8+4=12 \Omega
\end{aligned}\)
According to \(\mathrm{KCL}\), the current will get divided into two parts \(I_1\) and \(I_2\)
\(\therefore \quad \mathrm{I}_1=\frac{\mathrm{R}_2 \mathrm{I}}{\left(\mathrm{R}_1+\mathrm{R}_2\right)}\)
Substituting the values,
\(\begin{aligned}
& I_1=\frac{12}{9+12} \times 5 \\
& I_1=2.85 \mathrm{~A}
\end{aligned}\)
Potential difference between \(P\) and \(Q\) is
\(\begin{aligned}
& V=I_1 R \\
& V=2.85 \times 6 \\
& V=17 V
\end{aligned}\)
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