How to solve questions on current of electricity for UTME
How to solve questions on current of electricity for UTME
Question 1
A working electric motor takes a current of 1.5A when the p.d. across it is 250V. If its efficiency is 80%, the power output is{UTME 2001}
A. 469.0W B. 300.0W C. 4.8W D. 133.0W
Solution
Efficiency = (power output / power input)%
power input = IV = 1.5*250 = 375W
80/100 = power output / 375
power output = 80*375 / 100 = 300 W
B is the correct option
Question 2
A bread toaster uses a current of 4A when plugged in a 240 volts line. It takes one minute to toast slices of bread. What is the energy consumed by the toaster?{UTME 2001}
A. 3.60 x 103J B. 5.76 x 104J C. 1.60 x 102J D. 1.60 x 104J
Solution
one minute is 60 seconds
Energy consumed = IVt = 4*240*60 = 57600 J = 5.76 x 104 J
B is the correct answer
Question 3
In the circuit diagram above, the ammeter reads a current of 3A when R is 5Ω and 6A when R is 2Ω. Determine the value of x.{UTME 2001}
A. 10Ω B. 8Ω C. 4Ω D. 2Ω
Solution
X and R are in parallel
Effective resistance = (X*R)/(R+X)
When current is 3A, the effective resistance = 5X /(5+X)
V = IR
V = 3[5X /(5+X)]
when current is 6A, the effective resistance = 2X /(2+X)
V = IR
V = 6[2X /(2+X)]
Since R and X are in parallel, the same voltage will be applied across the two resistors
3[5X /(5+X)] = 6[2X /(2+X)]
15X /(5+X) = 12X/(2+X)
15/(5+X) = 12/(2+X)
cross multiply
15(2+x) = 12(5+x)
30 +15x = 60 + 12x
3x = 30
x = 10Ω
A is the correct answer
Question 4
A cell of emf 1.5V is connected in series with a 1Ω resistor and a current of 0.3A flows through the resistor. Find the
internal resistance of the cell.{UTME 2013)
A. 1.0 Ω B. 4.0 Ω C. 3.0 Ω D. 1.5 Ω
Solution
E = I(R+r)
1.5 = 0.3(1 + r)
1 + r = 1.5/0.3
1 + r =5
r = 5 – 1 = 4Ω
B is the correct option
Question 5
Which of the following obeys ohms law?{UTME 2013}
A. Glass B. Glectrolytes C. Metals D. Diode
Solution
Ohm’s law is about metallic material only provided the temperature and all other physical factors rmain constant.
C is the correct option
Question 6
Six identical cells, each of e.m.f. 2V are connected as shown above. The effective e.m.f. of the cell is{UTME 2012}
A. 0V B. 4V C. 6V D. 12V
Solution
cells arranged in series will have effective emf of
E = E1 + E2 + E3 +…
E = 2+2+2+2+2+2 = 12 V
D is the correct option
Question 7
In the circuit above, three resistors, 2Ω, 4 Ω and 12 Ω are connected in parallel and a 12 V battery is connected across the combination. The current flowing through the 12 Ω resistor is{UTME 2011}
A. 9.6 A B 14.4 A C. 1.0 A D. 3.2 A.
Solution
The current flowing through the 12 Ω resistor will be, V =IR
Since they are all in parallel, they will have the same voltage across them,
12 = I*12
I = 12/12 = 1A
C is the correct option
Question 8
The diagram above shows a balanced metre bridge, the value of x is{UTME 2010}
A. 66.7 cm B. 25.0 cm C. 33.3 cm D. 75.0 cm.
Solution
R1 /R2 = L1 / L2
R1, the resistors are in parallel, so the effective resistance = 8*8/(8+8) = 64/16 =4Ω
4 / x = 8 / (100-x)
4(100-x) = 8x
400 -4x = 8x
12x = 400
x = 400 / 12 = 33.33 cm
C is the correct option
Question 9
In the diagram above, a 200 W bulb is lighted by a 240 V a.c mains supply. If 1kWh is sold at N 40, the cost of keeping the bulb lighted for a day is{UTME 2010}
A. N 192.00 B. N 1.92 C. N 19.20 D. N 1,920.00.
Solution
Energy consumed = power(KW) * time(hour)
Energy consumed = 200/1000 * (a day = 24hrs) = 0.2*24 = 4.8KWh
1kWh is sold at N 40
4.8KWh will be N40*4.8 = N192.00
A is the correct option
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