JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 3)
In the given circuit the the internal resistance of the 18 V cell is negligible. If R1 = 400 $$\Omega $$, R3 = 100 $$\Omega $$ and R4 = 500 $$\Omega $$ and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be :
_9th_January_Evening_Slot_en_3_1.png)
300 $$\Omega $$
450 $$\Omega $$
550 $$\Omega $$
230 $$\Omega $$
Explanation
Voltage accross resistance R4 = 5 V
$$ \therefore $$ IR4 = 5 V
$$ \Rightarrow $$ 500 $$ \times $$ I = 5
$$ \Rightarrow $$ I = $${1 \over {100}}$$ A
$$ \therefore $$ Voltage across resistor R3 = $${1 \over {100}}\left( {100} \right)$$ = 1 A
$$ \therefore $$ Total drop in resistance R3 and R4 = 5 + 1 = 6V
So, voltage accross R2 resistance is also 6V as R3, R4 and R2 are in parallel
$$ \therefore $$ Voltage accross R1 resistor R1 resistor = 18 $$-$$ 6 = 12 V
$$ \therefore $$ Current through R1 resistor = $${{12} \over {400}}$$ = $${3 \over {100}}$$ A
$$ \therefore $$ Current through R2 resistor
= $${3 \over {100}} - {1 \over {100}}$$
= $${2 \over {100}}$$ A
$$ \therefore $$ $$\left( {{2 \over {100}}} \right)$$ R2 = 6
$$ \Rightarrow $$ R2 = 300 $$\Omega $$
$$ \therefore $$ IR4 = 5 V
$$ \Rightarrow $$ 500 $$ \times $$ I = 5
$$ \Rightarrow $$ I = $${1 \over {100}}$$ A
$$ \therefore $$ Voltage across resistor R3 = $${1 \over {100}}\left( {100} \right)$$ = 1 A
$$ \therefore $$ Total drop in resistance R3 and R4 = 5 + 1 = 6V
So, voltage accross R2 resistance is also 6V as R3, R4 and R2 are in parallel
$$ \therefore $$ Voltage accross R1 resistor R1 resistor = 18 $$-$$ 6 = 12 V
$$ \therefore $$ Current through R1 resistor = $${{12} \over {400}}$$ = $${3 \over {100}}$$ A
$$ \therefore $$ Current through R2 resistor
= $${3 \over {100}} - {1 \over {100}}$$
= $${2 \over {100}}$$ A
$$ \therefore $$ $$\left( {{2 \over {100}}} \right)$$ R2 = 6
$$ \Rightarrow $$ R2 = 300 $$\Omega $$
Comments (0)
