KCET · Physics · Current Electricity
Three conductors draw currents of \(1 \mathrm{~A}, 2 \mathrm{~A}\) and 3 A respectively, when connected in turn across a battery. If they are connected in series and the combination is connected across the same battery, the current drawn will be
- A \(\frac{2}{7} \mathrm{~A}\)
- B \(\frac{3}{7} \mathrm{~A}\)
- C \(\frac{4}{7} \mathrm{~A}\)
- D \(\frac{5}{7} \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{7} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Let the three conductors are \(R_{1}, R_{2}\) and \(R_{3}\) and the current drawn by them \(1 \mathrm{~A}, 2 \mathrm{~A}\) and \(3 \mathrm{~A}\) respectively when connected in turn across a battery.
\(\therefore V=R_{1}, \quad V=2 R_{2}\) and \(V=3 R_{3}\)
So, \(\quad R_{1}=V, R_{2}=\frac{V}{2}\) and \(R_{3}=\frac{V}{3}\)
When the conductors are connected in series, across the same battery, then
\(\Rightarrow V=l\left[R_{1}+R_{2}+R_{3}\right] \)
\( \Rightarrow V=I\left[V+\frac{V}{2}+\frac{V}{3}\right] \)
\( 1=I\left[\frac{6+3+2}{6}\right] \)
\( \text {or } I=\frac{6}{11} \mathrm{~A}\)
\(\therefore V=R_{1}, \quad V=2 R_{2}\) and \(V=3 R_{3}\)
So, \(\quad R_{1}=V, R_{2}=\frac{V}{2}\) and \(R_{3}=\frac{V}{3}\)
When the conductors are connected in series, across the same battery, then
\(\Rightarrow V=l\left[R_{1}+R_{2}+R_{3}\right] \)
\( \Rightarrow V=I\left[V+\frac{V}{2}+\frac{V}{3}\right] \)
\( 1=I\left[\frac{6+3+2}{6}\right] \)
\( \text {or } I=\frac{6}{11} \mathrm{~A}\)
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