Question

In Fig 27-66, the ideal battery has emf30 V, the resistances are${\mathit{R}}_{\mathbf{1}}\mathbf{=}$20 kΩ,and${\mathit{R}}_{\mathbf{2}}\mathbf{=}$10 kΩ, and the capacitor is uncharged. When the switch is closed at time t= 0, what is the current in

(a) Resistance 1 and

(b) Resistance 2?

(c) A long time later, what is the current in resistance 2?

Short answer

(a)The current in resistance $R_1$is$i_1=1.5\times {10}^{-3}A$

(b)The current in resistance i$R_2$s$i_2=0$

(c)Current in resistance$R_2$after long time later is$i=1.0\times {10}^{-3}A$

Step-by-step solution

Determine the given quantities

Consider the given values of the resistances and emf is:

$$\begin{gathered} \text{ε}=30\text{V} \\ {\text{R}}_1=20k\Omega \\ {R}_2=10k\Omega \end{gathered}$$

Determine the concept of Ohm’s law

According to Ohm’s law, the direct current flowing in a conductor is directly proportional to the potential difference between its ends.

Formulae:

$$\begin{gathered} I=\frac{V}{R} \\ {R}_{eq}=R_1+R_2 \end{gathered}$$

Step 3:(a) Determine current in resistor R1

Initially,thecapacitor is uncharged, sothevoltage across resistancewould be zero, and the voltage across$R_1$is 30 V.

By Ohm’s law:

Current in resistance$R_1$

$\begin{array}{c}{i}_1=\frac{30}{20\times {10}^3}\\ =1.5\times {10}^{-3}A\end{array}$

The current in resistance$R_1$ is$i_1=1.5\times {10}^{-3}A$

Step 4:(b) Determine of current in resistor R2

Current in resistance $R_2$is $i_2=0$this is because the voltage drop across this resistance is 0.

Step 5:(c) Determine the current in resistor R2 after long time later

A long time later, capacitor reduces to zero, so resistance $R_1$and$R_2$will be in series.

Hence,

$\begin{array}{c}{R}_{eq}=R_1+R_2\\ =(20\times {10}^3)+(10\times {10}^3)\\ =30\times {10}^3\Omega \end{array}$

Solve for the value of the current as:

$\begin{array}{c}i=\frac{30}{30\times {10}^3}\\ =1.0\times {10}^{-3}A\end{array}$

Current in resistance $R_2$after long time later is$i=1.0\times {10}^{-3}A$