JEE Mains · Physics · STD 12 - 3. current electricity
In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given \(R_1 = 15\,\Omega \)\(,R _2 = 10\,\Omega ,\)\( R_3 = 20\,\Omega ,\)\( R_4 = 5\,\Omega ,\)\(R_5 = 25\,\Omega ,\)\(R_6 = 30\,\Omega , \)\(E = 15\,V\)
- A \(13/24\)
- B \(7/18\)
- C \(9/32\)
- D \(20/3\)
Answer & Solution
Correct Answer
(C) \(9/32\)
Step-by-step Solution
Detailed explanation
\(R_{e q}=15+\frac{25}{3}+30=\frac{45+25+90}{3}=\frac{160}{3}\) \(\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eq}}}=\frac{15 \times 3}{160}=\frac{9}{32} \,\mathrm{amp}\).
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