Question

For an ideal battery (r=0?), closing the switch in the figure (Figure 1) does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 2.0 V battery has an internal resistance 0.2 ? and that the resistance of a glowing bulb is 8.0 ? .

Part A

What is the current through bulb A when the switch is open?

Part B

What is the current through bulb A after the switch has closed?

Part C

By what percent does the current through A change when the switch is closed?

Answer 1

A) Apply Kirchhoff's loop rule to find the current. When the switch is open, the bulb \(\mathrm{B}\) is not connected. So, \(\varepsilon-I r-I R_{A}=0\)

Here, I is the current in the circuit, \(\mathrm{R}_{\mathrm{A}}\) is the resistance of the bulb \(\mathrm{A}\). \(\varepsilon=I\left(r+R_{\mathrm{A}}\right)\)

\(I=\frac{\varepsilon}{\left(r+R_{\mathrm{A}}\right)}=\frac{2.0 \mathrm{~V}}{0.2 \Omega+8.0 \Omega}\)

\(I=0.244 \mathrm{~A}\)

B) When the switch is closed, then the two bulbs are connected in parallel. So, the current will be divided. Let the current across the bulb \(\mathrm{A}\) be \(I_{1}\) and bulb \(\mathrm{B}\) be \(I_{1} .\) However, from Kirchhoff's first law, \(I=I_{1}+I_{2}\)

Apply Kirchhoff' loop rule to the left loop. \(\varepsilon-I r-I_{1} R_{\mathrm{A}}=0\)

\(\varepsilon-\left(I_{1}+I_{2}\right) r-I_{1} R_{\mathrm{A}}=0\)

\(\varepsilon-I_{2} r-I_{1} r-I_{1} R_{\mathrm{A}}=0\)

\(2.0 \mathrm{~V}-I_{1}(0.2 \Omega+8.0 \Omega)-(0.2 \Omega) I_{2}\)

\((8.2 \Omega) I_{1}+(0.2 \Omega) I_{2}=2.0 \mathrm{~V}\)

Apply Kirchhoff's loop rule to the right loop. \(I_{1} R_{\mathrm{A}}-I_{2} R_{\mathrm{B}}=0\)

\(I_{2}=\frac{I_{1} R_{\mathrm{A}}}{R_{\mathrm{B}}}=\frac{I_{1}(8.0 \Omega)}{8.0 \Omega}=I_{1} \quad\) Replace it in (1)

\((8.2 \Omega) I_{1}+(0.2 \Omega) I_{1}=2.0 \mathrm{~V}\)

\(I_{1}=\frac{2.0 \mathrm{~V}}{(8.2 \Omega)+(0.2 \Omega)}\)

\(I_{1}=0.238 \mathrm{~A}\)

Percentage change:

\(\%\) change of current through the bulb \(\mathrm{A}\) is

\(\%\) change \(=\frac{0.238 \mathrm{~A}-0.244 \mathrm{~A}}{0.244 \mathrm{~A}} \times 100\)

\(\%\) change \(=2.46 \%\)