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Two resistors of 4⁢Ω and 6⁢Ω are connected in parallel to a cell to draw a current of 0.5 A from the cell. (i) Draw a labelled diagram showing the above arrangement.

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Question

Two resistors of \(4 \Omega\) and \(6 \Omega\) are connected in parallel to a cell to draw a current of 0.5 A from the cell.

  1. Draw a labelled diagram showing the above arrangement.
  2. Calculate the current in each resistor.
Numerical
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Solution

(i) Connect both resistors across the same two terminals of the cell; the ammeter measures the total current in series with the source:

(ii) Given: \(R_1=4\,\Omega\)

\(R_2=6\,\Omega\)

\(I=0.5\,A\)

The equivalent resistance is:

\[ R=\frac{R_1R_2}{R_1+R_2}\]

\[=\frac{4\times6}{4+6}\]

\[=2.4\ \Omega\]

The potential difference across the parallel pair is:

\[V=IR\]

\[=0.5\times2.4\]

\[=1.2\ \mathrm{V}\]

Current through the 4 Ω resistor:

\[ I_1=\frac{V}{R_1} \]

\[=\frac{1.2}{4}\]

\[=0.3\ \mathrm{A}\]

Current through the 6 Ω resistor:

\[ I_2=\frac{V}{R_2}\]

\[=\frac{1.2}{6}\]

\[=0.2\ \mathrm{A}\]

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Chapter 4: Electricity and Magnetism - Frequently Asked Questions in Board Examination [Page 286]

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Frank Physics Part 2 [English] Class 10 ICSE
Chapter 4 Electricity and Magnetism
Frequently Asked Questions in Board Examination | Q 36. (a) | Page 286
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