The circuit shown in the figure consists of a battery emf ε=10 V ; a capacitor of capacitance C=1.0 μ F three resistors of values R1=2 Ω, R2=2 Ω and R3=1 Initially the capacitor is completely uncharged and switch S is open. The switch S is closed at t=0 . Then

Question

The circuit shown in the figure consists of a battery emf ε=10 V ; a capacitor of capacitance C=1.0 μ F three resistors of values R1=2 Ω, R2=2 Ω and R3=1 Initially the capacitor is completely uncharged and switch S is open. The switch S is closed at t=0 . Then (A) the current through resistor R3 at the moment switch closed is zero (B) the current through resistor R3 a long time after switch closed is 5 A (C) the ratio of current through R1 and R2 is alw constant (D) all of these

Tardigrade Admin · Accepted Answer

At t=∞, the equivalent circuit is I=(ε/R3+(R1 R2)R1+R2) =(10/1+((2)(2))2+2)=(10/1+1)=5 A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f51813d9a823cdf906a7e7ecb6cb15f2-.png" /> Also, IR1 R1=IR2 R2 ⇒ (IR1/IR2)=(R2/R1)= constant

Q. The circuit shown in the figure consists of a battery emf $ε = 10 V;$ a capacitor of capacitance $C = 1.0 μ F$ three resistors of values $R_{1} = 2 Ω, R_{2} = 2 Ω$ and $R_{3} = 1$ Initially the capacitor is completely uncharged and switch $S$ is open. The switch $S$ is closed at $t = 0.$ Then

the current through resistor $R_{3}$ at the moment switch closed is zero

the current through resistor $R_{3}$ a long time after switch closed is $5 A$

the ratio of current through $R_{1}$ and $R_{2}$ is alw constant

all of these

At $t = \infty$ , the equivalent circuit is
$I = \frac{ε}{R _{3} + \frac{R _{1} R _{2}}{R _{1} + R _{2}}}$
$= \frac{10}{1 + \frac{( 2 ) ( 2 )}{2 + 2}} = \frac{10}{1 + 1} = 5 A$

Also, $I_{R_{1}} R_{1} = I_{R_{2}} R_{2}$
$\Rightarrow \frac{I _{R_{1}}}{I _{R_{2}}} = \frac{R _{2}}{R _{1}} =$ constant

Q. The circuit shown in the figure consists of a battery emf ε=10V; a capacitor of capacitance C=1.0μF three resistors of values R1​=2Ω,R2​=2Ω and R3​=1 Initially the capacitor is completely uncharged and switch S is open. The switch S is closed at t=0. Then

the current through resistor R3​ at the moment switch closed is zero

the current through resistor R3​ a long time after switch closed is 5A

the ratio of current through R1​ and R2​ is alw constant

all of these

At t=∞, the equivalent circuit is I=R3​+R1​+R2​R1​R2​​ε​ =1+2+2(2)(2)​10​=1+110​=5A Also, IR1​​R1​=IR2​​R2​ ⇒IR2​​IR1​​​=R1​R2​​= constant

the current through resistor $R_{3}$ at the moment switch closed is zero

the current through resistor $R_{3}$ a long time after switch closed is $5 A$

the ratio of current through $R_{1}$ and $R_{2}$ is alw constant

At $t = \infty$ , the equivalent circuit is
$I = \frac{ε}{R _{3} + \frac{R _{1} R _{2}}{R _{1} + R _{2}}}$
$= \frac{10}{1 + \frac{( 2 ) ( 2 )}{2 + 2}} = \frac{10}{1 + 1} = 5 A$

Also, $I_{R_{1}} R_{1} = I_{R_{2}} R_{2}$
$\Rightarrow \frac{I _{R_{1}}}{I _{R_{2}}} = \frac{R _{2}}{R _{1}} =$ constant