JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 14)
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15$$\Omega$$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.
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4.87 $$\mu$$A
4.87 mA
2.44 mA
2.44 $$\mu$$A
Explanation
$${{{V_B} - 10} \over {100}} + {{{V_B} - {V_D}} \over {15}} + {{{V_B} - 0} \over {10}} = 10$$
$${{{V_B} - 10} \over {20}} + {{{V_B} - {V_D}} \over 3} + {{{V_B}} \over 2} = 0$$
$$3{V_B} - 30 + 20{V_B} - 20{V_D} + 30{V_B} = 0$$
$$53{V_B} - 20{V_D} = 30$$ __________ (1)
Similarly,
$${{{V_D} - 10} \over {60}} + {{{V_D} - {V_B}} \over {15}} + {{{V_D} - 0} \over 5} = 0$$
$${V_D} - 10 + 4{V_D} - 4{V_B} + 12{V_D} = 0$$
$$ - 4{V_B} + 17{V_D} = 10$$ __________ (2)
after solving equation (1) & (2)
VD = 0.79 volt
VB = 0.86 volt
Then the current through the galvanometer
$$ = {{{V_B} - {V_D}} \over R}$$
$$ = {{0.86 - 0.79} \over {15}}$$ = 4.87 mA
$${{{V_B} - 10} \over {20}} + {{{V_B} - {V_D}} \over 3} + {{{V_B}} \over 2} = 0$$
$$3{V_B} - 30 + 20{V_B} - 20{V_D} + 30{V_B} = 0$$
$$53{V_B} - 20{V_D} = 30$$ __________ (1)
Similarly,
$${{{V_D} - 10} \over {60}} + {{{V_D} - {V_B}} \over {15}} + {{{V_D} - 0} \over 5} = 0$$
$${V_D} - 10 + 4{V_D} - 4{V_B} + 12{V_D} = 0$$
$$ - 4{V_B} + 17{V_D} = 10$$ __________ (2)
after solving equation (1) & (2)
VD = 0.79 volt
VB = 0.86 volt
Then the current through the galvanometer
$$ = {{{V_B} - {V_D}} \over R}$$
$$ = {{0.86 - 0.79} \over {15}}$$ = 4.87 mA
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