Institutt for fysikk, NTNU TFY4155/FY1303 Elektrisitet og magnetisme Vår 2004

Solution to øving 11

Guidance Monday March 22

Exercise 1

a) At first, we should try to realize that what we have here is the following circuit: [a parallel connection of R_{1}, R_{2} and R_{3}] coupled in series with [a parallel connection of R_{4} and R_{0} = 0] in series with [R_{5}]. In other words, the resistance R_{4} is ”cut short”, so that no current passes through R_{4}. (Alternatively: We have the same value for the potential on each side of R_{4}. Then, no current can pass through it.) Thus, the total resistance is

R = (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} )^{−1} + R_{5}

b) It should be clear that the total current I in the circuit must be the same as the current I_{5} passing through R_{5}. Further, it should also be clear that I must distribute itself on the three currents passing through R_{1}, R_{2} and R_{3}: I = I_{1} + I_{2} + I_{3}. In a) above, we have already concluded that no current passes through R_{4}: I_{4} = 0. The voltage drop across the three upper resistances is the same:

V′ = R_{1}I_{1} = R_{2}I_{2} = R_{3}I_{3}

The voltage drop across R_{5} is

V′′ = R_{5}I_{5} = R_{5}I = R_{5} \frac{E}{R}

These two together must equal the value of the voltage source:

E = V′ + V′′

Thus

V′ = E − V′′ = E − R_{5} \frac{E}{R} = E (1 − \frac{R_{5}}{R} )

And finally,

I_{1} = \frac{V′}{R_{1}}

I_{2} = \frac{V′}{R_{2}}

I_{3} = \frac{V′}{R_{3}}

c) With the given numerical values, we have

R = (1 + \frac{1}{2} + \frac{1}{3})^{−1} + 5 = \frac{61}{11} Ω

Thus,

V′ = 9 · (1 − \frac{5}{61/11}) = \frac{54}{61} V

and

V′′ = 9 − \frac{54}{61} = \frac{495}{61} V

The various currents are

I_{1} = \frac{54}{61 · 1} = \frac{54}{61} A ≃ 0.885 A

I_{2} = \frac{54}{61 · 2} = \frac{27}{61} A ≃ 0.443 A

I_{3} = \frac{54}{61 · 3} = \frac{18}{61} A ≃ 0.295 A

I_{5} = I = \frac{9 · 11}{61} = \frac{99}{61} A ≃ 1.623 A

Exercise 2

a) Bulb 1 will be brightest in circuit B and weakest in circuit A. In circuit A, the voltage drop across bulb 1 is only 1/3 of the emf of the applied voltage source. In B, the voltage drop across all the bulbs is equal to the emf of the applied voltage source. In C, the parallel connection of 2 and 3 constitutes a resistance (1/R + 1/R)^{−1} = R/2, if the resistance in one bulb is R. Then, the voltage drop over bulb 1 becomes 2/3 of the emf of the applied voltage source, and the light intensity somewhere between that in A and B.

b) In A, we end up with an open (”broken”) circuit, and therefore zero current, i.e., bulbs 1 (and 2) go out. In B, bulb 3 does not at all influence the voltage across bulb 1, so the light intensity is unchanged. In C, we end up with two resistances R coupled in series, so the voltage drop across bulb 1 must be half of the emf of the applied voltage source. I.e., smaller than what it was when bulb 3 was in place. Thus, the light intensity becomes smaller in circuit C.

Exercise 3

Let us put time t = 0 when the bullet breaks the circuit in point A. Before this, we have a stationary situation, with voltage drop

V_{0} = V_{R} = V_{C}

over both the resistance R and the capacitance C. At t = 0, the circuit is broken (”opened”) in A, and we are left with the loop with R and C (and no voltage source anymore). As derived in the lectures, we now get a discharge of the capacitor, with a time dependence

Q(t) = Q_{0}e^{−t/RC}

where

Q_{0} = Q(0) = CV_{0}

After a certain time t_{1}, the circuit is broken at B, which means that the discharge stops. Then we have a charge

Q(t_{1}) = Q_{0}e^{−t_{1}/RC}

on the capacitor. This corresponds to a voltage drop

V (t_{1}) = \frac{Q(t_{1})}{C} = \frac{Q_{0}}{C} e^{−t_{1}/RC}

across the capacitor. During this time, the bullet has moved a distance d, so that its velocity must be

v = d/t_{1}

We solve for t_{1} in the expression for V (t_{1}):

t_{1} = RC ln \frac{V_{0}}{V (t_{1})}

where we have used Q_{0} = V_{0}C and ln x = − ln(1/x). With numerical values inserted:

t_{1} = 250 · 10^{−6} · ln \frac{9}{4} = 2.03 · 10^{−4} s

The velocity becomes

v = d/t_{1} = 0.1/2.03 · 10^{−4} = 493 m/s