KCET · Physics · Magnetic Effects of Current
Two resistors of resistances \( 2 \Omega \) and \( 6 \Omega \) are connected in parallel. This combination is then connected to a battery of emf \( 2 \mathrm{~V} \) and internal resistance \( 0.5 \Omega \). What is the current flowing through the battery?
- A (1)
- B \( \frac{4}{3} A \)
- C \( \frac{4}{17} \mathrm{~A} \)
- D \( 1 A \)
Answer & Solution
Correct Answer
(D) \( 1 A \)
Step-by-step Solution
Detailed explanation
Current flowing through circuit is \( I=\frac{E}{R_{e q}+r} \)
Since resistance are connected in parallel, therefore,
\( \frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow \frac{1}{R_{e q}}=\frac{1}{2}+\frac{1}{6}=\frac{3+1}{6}=\frac{4}{6} \)
\( \Rightarrow R_{\text {애 }}=\frac{6}{4}=\frac{3}{2}=1.5 \Omega \)
Therefore, \( I=\frac{2}{1.5+0.5}=\frac{2}{2}=1 \mathrm{~A} \)
Thus, current flowing through the battery is \( 1 \mathrm{~A} \)
Since resistance are connected in parallel, therefore,
\( \frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow \frac{1}{R_{e q}}=\frac{1}{2}+\frac{1}{6}=\frac{3+1}{6}=\frac{4}{6} \)
\( \Rightarrow R_{\text {애 }}=\frac{6}{4}=\frac{3}{2}=1.5 \Omega \)
Therefore, \( I=\frac{2}{1.5+0.5}=\frac{2}{2}=1 \mathrm{~A} \)
Thus, current flowing through the battery is \( 1 \mathrm{~A} \)
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