Single Loop Circuits (emf, terminal P.d, internal resistance) Problems and Solutions 2

 Problem #1

(a) In Fig. 1, what value must R have if the current in the circuit is to be 1.0 mA? Take ε₁ = 2.0 V, ε₂ = 3.0 V, and r₁ = r₂ = 3.0 Ω. (b) What is the rate at which thermal energy appears in R?

Fig.1

Answer:
Known:
current in the circuit, i = 1.0 mA
emf, ε₁ = 2.0 V
emf, ε₂ = 3.0 V
internal resistance, r₁ = r₂ = 1.0 Ω

(a) value must R have if the current in the circuit is to be 1.0 mA is

Assuming the direction of current is counterclockwise and using the loop rule in the counterclockwise direction:

−ε₁ − ir₂ + ε₂ − ir₁ − iR = 0

i = (ε₂ – ε₁)/(r₂ + r₁ + R)

i = (3.0 V ─ 2.0 V)/(1 Ω + 1 Ω + R)

0.001 = 1/(2 + R)

R = 998 Ω

(b) the rate of thermal energy in R is

 P = i²R = (0.001A)²(998 Ω) = 998 μW

Problem #2
A 10-km-long underground cable extends east to west and consists of two parallel wires, each of which has resistance 13 Ω/km. An electrical short develops at distance x from the west end when a conducting path of resistance R connects the wires (Fig. 2). The resistance of the wires and the short is then 100 Ω when measured from the east end and 200 Ω when measured from the west end. What are (a) x and (b) R?

Fig.2

Answer:
Known:
Resistance, R = 13 Ω/km
cable length, L = 10 km

(a) Let R₁ be the resistance on the west side and R₂ on the east.

R₁ + R₂ = 13Ω/km x 10 km = 130 Ω

R₂ = 130 – R₁

R₁ + R + R₁ = 2R₁ + R = 200 Ω

R₂ + R + R₂ = 2R₂ + R = 100 Ω

Substitute the expression for R₂ using R₁

2(130 – R₁) + R = 100

260 – 100 = 2R₁ – R = 160

But,  2R₁ + R = 200

Add the two equations

4R₁ = 360

R₁ = 90 Ω,

R₂ = 130 - 90 = 40 Ω, then

x = (90 Ω/130 Ω) x 10 km = 6.9 km

(b) R = 200 – 2(90 Ω) = 20 Ω

Problem #3
A solar cell generates a potential difference of 0.10 V when a 500 Ω resistor is connected across it, and a potential difference of 0.15 V when a 1000 Ω resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is 5.0 cm², and the rate per unit area at which it receives energy from light is 2.0 mW/cm². What is the efficiency of the cell for converting light energy to thermal energy in the 1000 Ω external resistor?

Answer:
Known:
area of the cell is A = 5.0 cm²
potential difference when a R₁ = 500 Ω resistor is V₁ = 0.10 V
potential difference when a R₂ = 1000 Ω resistor is V₂ = 0.15 V

Fig.3

(a) internal resistance, r

For the first circuit:

V₁ = ε ─ i₁r

V₁ =  ε ─ (V₁/R₁)r

0.10 V = ε ─ (0.10 V/500 Ω)r

or 10 = 100ε ─ (10r/500)  (*)

For the second circuit:

V₂ = ε ─ i₂r

V₂ =  ε ─ (V₂/R₂)r

0.15 V = ε ─ (0.15 V/1000 Ω)r

or 15 = 100ε ─ (15r/1000)  (**)

from (*) and (**),

5 = ─(15r/1000) + (20r/1000)

r = 1000 A

(b) emf of the solar cell is

V₁ = ε ─ i₁r

0.10 V = ε ─ (0.10 V/500 Ω) x 1000 A

ε = 0.3 V

(c) the efficiency of the cell is

receives energy from light is P_rec = 2.0 mW/cm² = 0.002 W/cm²

eff = P/(P_rec.A)

with P = V²/r, then

eff = V²/(P_rec.Ar)

eff = (0.15 V)²/(0.002 W/cm² x 5 cm² x 1000 Ω) = 0.00225 or 0.25%

Problem #4
In Fig. 4, battery 1 has emf ε₁ = 12.0 V and internal resistance r₁ = 0.016 Ω and battery 2 has emf ε₂ = 12.0 V and internal resistance r₂ = 0.012 Ω. The batteries are connected in series with an external resistance R. (a) What R value makes the terminal-to-terminal potential difference of one of the batteries zero? (b) Which battery is that?

Fig.4

Answer:

Known:
emfs ε₁ = 12.0 V
emfs ε₂ = 12.0 V
resistor 1, r₁ = 0.016 Ω
resistor 1, r₂ = 0.012 Ω

(a) R value makes the terminal-to-terminal potential difference of one of the batteries zero.

The direction of current in this circuit is clockwise, using the loop rule in the clockwise direction:

ε₁ − ir₂ + ε₂ − ir₁ − iR = 0 (*)

If the terminal-to-terminal potential difference of one battery is zero, so we have:

ε_x ─ ir_X = 0 (**)

and

ε_y ─ ir_y = iR

R = (ε_y/i) ─ r_y

X can be 1 or 2, so Y can be 2 or 1 accordingly

So, R = (ε_y/i) ─ r_y =  (ε_y/ε_x)r_x ─ r_y

R = r_x ─ r_y  = r₁ ─ r₂ = 0.016 Ω ─ 0.012 Ω = 4.0 mΩ

(b) battery 1  

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