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GUJCET · Physics · Current Electricity
The total current supplied to the given circuit by battery is ________.
- A 6 A
- B 4 A
- C 9 A
- D 2 A
Answer & Solution
Correct Answer
(A) 6 A
Step-by-step Solution
Detailed explanation
(A)
\(6 \Omega\) and \(2 \Omega\) are in parallel
\(\begin{array}{l}R^{\prime}=\frac{6 \times 2}{6+2} \\R^{\prime}=1.5 \Omega\end{array}\)
\(1.5 \Omega\) and \(R ^{\prime}\) are in series
\(\begin{array}{l}R^{\prime \prime}=1.5+R^{\prime} \\R^{\prime \prime}=1.5+1.5 \\R^{\prime \prime}=3 \Omega\end{array}\)
\(3 \Omega\) and \(R ^{\prime \prime}\) are in parallel
\(\begin{array}{l}R=\frac{3 \times 3}{3+3} \\R=\frac{9}{6} \\R=1.5 \Omega\end{array}\)
Current passing through battery
\(\begin{aligned} \therefore \quad I =\frac{ V }{ R } \\ I =\frac{9}{1.5} \\ I =6 A\end{aligned}\)
\(6 \Omega\) and \(2 \Omega\) are in parallel
\(\begin{array}{l}R^{\prime}=\frac{6 \times 2}{6+2} \\R^{\prime}=1.5 \Omega\end{array}\)
\(1.5 \Omega\) and \(R ^{\prime}\) are in series
\(\begin{array}{l}R^{\prime \prime}=1.5+R^{\prime} \\R^{\prime \prime}=1.5+1.5 \\R^{\prime \prime}=3 \Omega\end{array}\)
\(3 \Omega\) and \(R ^{\prime \prime}\) are in parallel
\(\begin{array}{l}R=\frac{3 \times 3}{3+3} \\R=\frac{9}{6} \\R=1.5 \Omega\end{array}\)
Current passing through battery
\(\begin{aligned} \therefore \quad I =\frac{ V }{ R } \\ I =\frac{9}{1.5} \\ I =6 A\end{aligned}\)
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