A cell of constant emf first connected to a resistance R_{1} and then connected to resistance R_{2}. If power delivered in both case is same then the internal resistance of the cell is :
We know, Power delivered = i^{2}R where i = current and R = Resistance
Power delivered in both cases is equal then
The seals of a galvanometer of resistance 100Ω contains 25 division. It gives a deflection of one division of passing a current of 4 × 10^{–4}A. The resistance in ohm to be added to it, so, that it may become a voltmeter of range 2.5 V is :
The deflection of one division of galvanometer is given by the current = 4 × 10^{–4} A
or the current sensitivity of galvanometer is
= 4 × 10^{–4} A/div
i_{g} = full scale deflection
= current sensitivity × total number of turns
= 4 × 10^{–4} × 25 = 10^{–2} A
Resistance of galvanometer = 100Ω
The value of resistance required,
The correct answer is: 150
The resistance of the filament of a lamp increase with the increase in temperature. A lamp rated 100W and 200V is connected across 220V power supply. If the voltage drops by 10% then the power of the lamp will be :
Let the resistance of the lamp filament be R. Then when the voltage drops expected power is
Here R' will be less than R, because now the rise in temperature will be less. Therefore P is more than
But it will not be 90% of earlier value, because fall in temperature is smaller.
The correct answers are: 81 W, between 81 and 90 W
The figure shows the variation of v with I at temperature T_{1} and T_{2}. (T_{1} – T_{2}) is proportional to :
R = R_{0} + R_{0}α(T_{1}T_{0})
tan3α = R_{0} + R_{0}α(T_{1}T_{0})...............(1)
tan2α = R_{0} + R_{0}α(T_{2}T_{0})...............(2)
tan3α  tan2α = R_{0}α(T_{1}T_{2})
(T_{1}T_{2}) α tan(3α)  tan(2α)
tan(3α2α) = (tan3α  tan2α)/(1 + tan3αtan2α)
(T_{1}  T_{2}) α tanα(1 + tan3αtan2α)
So, tanα is the right answer.
The potential difference across the terminal of a battery is 50 V when 11 A current is drawn and 60V when 1A current is drawn. The emf and the internal resistance of the battery are :
For a closed circuit cells supplies a constant current in the circuit.
Equation of cells E = V + ir
For V = 50 volt
E = 50 + 11r ...(i)
Similarly for V = 60V
E = 60 + r ...(ii)
From Equation (i) and (ii), we get
r = 1Ω
Substituting the value of r is equation (i), we get
E = 61V
The correct answer is: 61V, 1Ω
Masses of three wire of copper are in ratio 1:3:5, and their length are in ratio 5:3:1. The ratio of their electrical resistance is
Ratio of masses of wires
(m_{1}, m_{2}, m_{3}) = 1 : 3 : 5
Ratio of length of wires
Electrical resistance is given by
The correct answer is: 125 : 15 : 1
When a galvanometer is shorted by resistance S, its current capacity increase n times. If the same galvanometer is shunted by another resistance S', its current capacity will increased by n' is given by :
Using the relation
A wire of resistance 5Ω is drown out so that its new length is 3 times its original length. What is the resistance of the new wire?
Let L_{1} is the original length and L_{2} is the length after drawn out.
Let L_{1} = l and L_{2} – 3L_{1} = 3l
Volume is constant
The resistance of a wire at 300K is found to be 0.3Ω. If the temperature coefficient of resistance of wire is 1.5 × 10^{–3} K^{–1}, the temperature at which resistance become 0.6Ω is :
An electric immersion heater of 1.08KW is immersed in water. After the water has reached a temperature of 100°C, how much time will be required to produced 100g of steam?
L is the latent heat of vaporisation of water, the heat required to produced 1g of steam.
L = 540 cal = 540 × 4. 2 = 2268J.
Energy supplied = 1080 Js^{–1}
Time required to boil 100g of water
The correct answer is: 210 s
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