Question

For an ideal battery (r = 0 Ω), closing the switch in (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 1.50 V battery has an internal resistance r = 0.90 Ω and that the resistance of a glowing bulb is R = 8.00 Ω.

1. What is the current through bulb A when the switch is open? ANS: 0.17A

2. What is the current through bulb A after the switch has closed? ANS: 0.15A

3. By what percent does the current through A change when the switch is closed? ANS: -9.2%

I have the answers for all of the problems and I know how to solve the first two, but I have no idea how to solve the third one. Thanks!

Answer 1

Given is :-

Battery = 1.50 V

internal resistance r=0.90 ohms

and R=8.00 ohms

Now,

Part-1

When the swithc is open the internal resistance r and the resistance of bulb A are in series connection therefore the total combined resistance will be

$R_{total} = 0.90+8 = 8.90 \Omega$

Therefore the current through bulb A when the switch is open is

$I = \frac{V}{R_{total}}$

$I = \frac{1.50}{8.90}$

$\boxed{I=0.1685 \approx 0.17A}$

Part-2

When swtich is closed , both the bulb is parallel combination therefore their combined reistance is

$R' = \frac{R \times R}{R+ R}= \frac{R}{2}= \frac{8}{2}= 4 \Omega$

and the internal reistance is in series connection to this combined resistance therefore the total resistance of the ckt is

$R_{eq} = 4 +0.90 = 4.90 \Omega$

Now the current drawn from the battery is

$I = \frac{V}{R_{eq}}$

$I = \frac{1.5}{4.90}$

The current through bulb A when the switch is closed is

$I_a = \frac{1}{2}I = \frac{1}{2}(0.306)$

$\boxed{I_a =0.153 \approx 0.15 A}$

Part-3

% change in the curren is

$=\frac{0.153-0.1685}{0.1685} \times 100$

which gives us

$\boxed{-9.19 \approx -9.2 \%}$