When a 12 V battery is connected
across an unknown resistor, there is a current of 2.5 mA in the circuit.
Find the value of the resistance of the resistor. Answer: Given, Potential difference, V = 12 V
Current (I) across the resistor = 2.5mA = 2.5 x 10 -3 = 0.0025 A
Resistance, R =?
We know, R = V/I = 12 V ÷ 0.0025 A = 4800
A piece of wire of resistance R is cut into five equal
parts. These parts are then connected in parallel. If the equivalent
resistance of this combination is R′, then the ratio R/R′ is
1/25
1/5
5
25
Answer: (d) 25
Explanation: The piece of wire having resistance equal to R is cut into five equal parts. Therefore, resistance of each part would be R/5.
When all parts are connected in parallel, the resistance of total resistance can be given as follows:
1/R' = 5 x (5/R) = 25/R
Or, R/R' = 25
Which of the following terms does not represent electrical power in a circuit?
I2R
IR2
VI
V2/R
Answer: (b) IR2Explanation: We know that Power (P) = VI
After substituting the value of V = IR in this we get
P = (IR) I = I x R x I = I2R, Thus P = I2R
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
100 W
75 W
50 W
25 W
Answer: (d) 25 W
Explanation: Potential difference, V = 220 V, Power, P = 100 W
Therefore, power consumption at 100 V =?
To solve this problem, first of all resistance of the bulb is to be calculated.
We know that P = V2 ÷ R
Or, 100 W = (220 V)2 ÷ R
Or, R = 48400 ÷ 100 = 484 Ω
Now, when the bulb is operated at 110 V, then power can be calculated as follows:
P = 1102 ÷ 484 = 12100 ÷ 484 = 25 W
Thus, bulb will consume power of 25W at 110V
Two conducting wires of the same material and of equal
lengths and equal diameters are first connected in series and then
parallel in a circuit across the same potential difference. The ratio of
heat produced in series and parallel combinations would be
1:2
2:1
1:4
4:1
Answer: (d) 4 : 1
Explanation: Let the potential difference = V,
Resistance of the wire = R Resistance when the given wires connected in series = Rs Resistance when the given wires connected in parallel = Rp Heat produced when the given wires connected in series = Hs Heat produced when the given wires connected in parallel = Hp
Thus, resistance Rs when the given two wires connected in series = R + R = 2R
Resistance Rp when the wires are connected in parallel can be calculated as follows:
1/Rp = 1/R + 1/R = 2/R
Or, Rp = R/2
We know, heat produced H = I2R t
Ratio of heat produced in two conditions:
Hs : Hp = 2R ÷ R/2 = 4 : 1
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: Voltmeter is connected into parallel to measure the potential difference between two points in a circuit.
A copper wire has diameter 0.5 mm and resistivity of
1.6 × 10–8 Ω m. What will be the length of this wire to make its
resistance 10 Ω? How much does the resistance change if the diameter is
doubled?
Answer: Given, Diameter of wire = 0.5 mm
Hence, radius = 0.25 mm = 0.00025 m
Resistivity, ρ = 1.6 x 10-8 Ω m
Resistance (R) = 10 Ω and length = ?
Resistance (R1) when diameter is doubled = ?
We know;
When diameter is doubled, radius becomes 0.0005 m
The values of current I flowing in a given resistor for
the corresponding values of potential difference V across the resistor
are given below
Plot a graph between V and I and calculate the resistance of that resistor.
Answer: The slope of the graph will give the value of resistance.
Let us consider two points A and B on the slope.
Draw two lines from B along X-axis and from A along Y-axis, which meets at point C
Now, BC = 10.2 V – 3.4 V = 6.8 V
AC = 3 – 1 = 2 ampere
When a 12 V battery is connected across an unknown
resistor, there is a current of 2.5 mA in the circuit. Find the value of
the resistance of the resistor.
Answer: Given, Potential difference, V = 12 V
Current (I) across the resistor = 2.5mA = 2.5 x 10 -3 = 0.0025 A
Resistance, R =?
We know; R = V/I
= 12 V ÷ 0.0025 A = 4800 Ω
A battery of 9 V is connected in series with resistors
of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current
would flow through the 12 Ω resistor?
Answer: Given, potential difference, V = 9 V
Resistance of resistors which are connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω respectively
Current through resistor having resistance equal to 12Ω =?
Total effective resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
We know; I = V/R
= 9 V ÷ 13.4 Ω = 0.671 A
Since, there is no division of electric current, in the circuit if
resistors are connected in series, thus, resistance through the
resistor having resistance equal to 12 Ω = 0.671 A
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