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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Current Electricity

BITSAT 2014 · Physics · Current Electricity

For the circuit (figure), the current is to be measured. The...

For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance RG=60.00Ω\mathrm{R}_{\mathrm{G}}=60.00 \Omega converted to an ammeter by a shunt resistance rs=0.02Ω\mathrm{r}_{\mathrm{s}}=0.02 \Omega. The value of the current is For the circuit (figure), the current is to be measured. The ammeter shown is a galvanometer with a resistance R_ G

Options

  1. A0.79A0.79 \mathrm{ A}
  2. B0.29A0.29 \mathrm{ A}
  3. C0.99A0.99 \mathrm{ A}
  4. D0.8A0.8 \mathrm{ A}

Correct answer

C. 0.99A0.99 \mathrm{ A}

Step-by-step solution

RG=60.00Ω,\mathrm{R}_{\mathrm{G}}=60.00 \Omega, shunt resistance, rs=0.02Ω\mathrm{r}_{\mathrm{s}}=0.02 \Omega Total resistance in the circuit is RG+3=63Ω\mathrm{R}_{\mathrm{G}}+3=63 \Omega Hence, I=3/63=0.048A\mathrm{I}=3 / 63=0.048 \mathrm{ A} Resistance of the galvanometer converted to an ammeter is, RGrsRG+rs=60Ω×0.02Ω(60+0.02)Ω=0.02ΩTotalresistanceinthecircuit\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}_{\mathrm{s}}}=\frac{60 \Omega \times 0.02 \Omega}{(60+0.02) \Omega}=0.02 \Omega Total resistance in the circuit=0.02+3=3.02ΩHence,=3.02 \Omega Hence,\mathrm{I}=3/3.02=0.99A=3 / 3.02=0.99 \mathrm{ A}
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