Question 1

An electric generator has an e.m.f. of 240V and an internal resistance of 1Ω. If the current supplied by the generator is 20A when the terminal voltage is 220V, find the ratio of the power supplied to the power dissipated.{UTME 2008}

A. 11 : 1 B. 1 : 11 C. 12 : 11 D. 11 : 12

Solution

power supplied = IE

p= 20 * 240 = 4800

power dissipated = IV

p = 20 * 220= 4400 watt

ratio = 4800 / 4400

12 : 11

A is the correct answer

Question 2

Find the effective resistance in the diagram above.{UTME 2008}

A. 6Ω B. 12Ω C. 16Ω D. 24Ω

solution

the arrangement is series because there is no branching of current

Rt = R1 + R2 + R3 + R4 + R5 + R6

Rt = 4 + 4 + 4 + 4 + 4 + 4

Rt = 24Ω

D is the correct option

Question 3 and 4 are from cambridge may/june 2016 p11

Question 3

In the circuit shown, X is a variable resistor whose resistance can be changed from 5.0 Ω to 500 Ω. The e.m.f. of the battery is 12.0 V. It has negligible internal resistance

What is the maximum range of values of potential difference across the output?

A 1.3 V to 11.1 V B 1.3 V to 12.0 V C 1.5 V to 11.1 V D 1.5 V to 12.0 V

Solution

Vout = (Rx / Rx + 40)* Vin

since Rx is a variable resistor, the vout will give a varying voltage

when Rx is 5Ω

Vout = (5 / 5 +40)* 12

Vout = 1.3 V

When Rx is 500 Ω

Vout =( 500 / 500 + 40)* 12

Vout = 11.1 V

so Vout will range from 1.3 v to 11.1 v

A is the correct option

Question 4

There is a current from P to R in the resistor network shown.

The potential difference (p.d.) between P and Q is 3 V.

The p.d. between Q and R is 6 V.

The p.d. between P and S is 5 V.

Which row in the table is correct?

Solution

potential difference between Q and S = p.d. between P and S is 5 V – potential difference (p.d.) between P and Q is 3 V.

potential difference between Q and S = PS – PQ

potential difference between Q and S = 5 – 3 = 2v

also,

potential difference between Q and S= The p.d. between Q and R is 6 V – potential difference between S and R

2 = 6 – SR

SR = 4 v

A is the correct option

Question 5

A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 0.25 Ω is connected in series with two identical resistors X and a resistor Y, as shown in Fig below

The resistance of each resistor X is 0.15 Ω and the resistance of resistor Y is 2.7 Ω.

(i) Show that the current in the circuit is 2.8 A.

(ii) Calculate the potential difference across the battery.

(iii) Each resistor X connected in the circuit above is made from a wire with a cross-sectional area of 2.5 mm2. The number of free electrons per unit volume in the wire is 8.5 × 10^29 m–3. Calculate the average drift speed of the electrons in X.{cambridge may/ june 2016 p22}

Solution

the resistors are arranged in series

Rt = R1 + R2 + R3

Rt = 0.15 + 2.7 + 0.15

Rt = 3Ω

E = IR + Ir

9 = I (Rt + r)

9 = I ( 3 + 0.25)

9 = 3.25 * I

I = 9 / 3.25

I = 2.8 A

ii.

potential difference across the battery = IRt

p.d = 2.8 * 3

p.d = 8.3 v

iii

I = nevA

n is the number of charge

e is the electronic charge

v is the drift speed

A is the area

I is the current

A = 2.5 mm^2 which in meters will be 2.5 x 10^-6 m^2

v = I / evA

v = 2.77 / (8.5 × 10^29 × 1.6 × 10^–19 × 2.5 × 10^–6)

v = 8.147 x 10^-6 ms-1