JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 10)
A galvanometer of resistance 100 $$\Omega $$ has 50 divisions on its scale and has sensitivitv of 20 $$\mu $$A/division. It is
to be converted to a voltmeter with three ranges of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so
is
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Explanation
20 × 50 × 10–6 = 10–3 Amp
$${V_1} = {2 \over {{{10}^{ - 3}}}} = 100 + {R_1}$$
$$1900 = {R_1}$$
$${V_2} = {{10} \over {{{10}^{ - 3}}}} = \left( {2000 + {R_2}} \right)$$
$${R_2} = 8000$$
$${V_3} = {{20} \over {{{10}^{ - 3}}}} = 10 \times {10^3} + {R_3} = 10 \times {10^3}{R_3}$$
$${V_1} = {2 \over {{{10}^{ - 3}}}} = 100 + {R_1}$$
$$1900 = {R_1}$$
$${V_2} = {{10} \over {{{10}^{ - 3}}}} = \left( {2000 + {R_2}} \right)$$
$${R_2} = 8000$$
$${V_3} = {{20} \over {{{10}^{ - 3}}}} = 10 \times {10^3} + {R_3} = 10 \times {10^3}{R_3}$$
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