Electricity Exemplar Problems Long Answer Questions Part 2 Class 10 Science

Measure potential difference when all the three resistors are connected in parallel.
Remove one of the resistors and measure the potential difference for remaining two resistors in parallel.
Remove the second resistor and measure the potential difference for the remaining resistor.
It is observed that the voltmeter shows same reading in all these conditions. This shows, that same potential difference exists across three resistors which are connected in parallel combination.

Question 34. What is Joule's heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Answer: The Joule's Law of Heating states that the heat produced in a resistor is:

Directly proportional to the square of current for a given resistor.
Directly proportional to resistance for a given current, and
Directly proportional to the time for which the current flows through the resistor.

This can be expressed by following equation:

`H=I^2` Rt

Where, I is electric current, R is resistance, t is time and H is the heating effect of electric current.

This can be demonstrated by a simple experiment, in which an immersion rod water heater can be used. This should be connected to a regulator which we use for regulating the speed of an electric fan.

In the first step, keep the knob of regulator at mark 1. Count the time taken by imersion rod to heat a mug of water.
In the second step, keep the knob of regulator at mark 2. Count the time taken by immersion rod to heat a mug of water.
Similarly, increase the magnitude of current and count the time taken by immersion rod to heat a mug of water.
It is observed that it takes less time for immersion rod to heat water when electric current is increased with the help of regulator.
Similarly, electic supply is increased in incremental value of time and temperature of water is measured. This shows that time has positive impact on heating effect of electric current.

Question 35. Find out the following in the electric circuit given in following figure.

(a) Effective resistance of two 8 Ω resistors in the combination

Answer: Let the total effective resistance in the combination = R

Therefore,`1/R= 1/(8Ω)+1/(8Ω)= (1+1)/(8 Ω)``=2/(8 Ω)=1/(4 Ω)`

Thus,`R=4Ω`

Therefore, effective resistance = 4Ω

(b) Current flowing through 4 Ω resistor

Answer: Given, potential difference through the circuit, V = 8V

The total effective resistance of the circuit, R = resistance in series + effective resistance in parallel combination

⇒R = 4Ω + 4Ω = 8Ω

Thus,electric current,I through the circuit`= V/R`

⇒`I= (8V)/(8Ω)=1A`

Since, there is no division of electric current in series combination,

thus, electric current through the 4Ω resistor = 1A

Answer: Since there is no division of electric current, in the case of series connection.

Thus, electric current through the 4Ω resistor = 1A

Since,potential difference,V across the 4Ω resistor=IR

Therefore,V= 1A×4Ω=4V

(d) Power dissipated in 4 Ω resistor

Answer: Electric current, I = 1A

Potential difference, through 4Ω resistor = 4V

Thus, power, P = V x I

⇒P = 4V x 1A = 4 watt

Thus, power dissipated in 4Ω resistor = 4W

(e) Difference in ammeter readings, if any.

Answer: Since resistors are connected in series, thus there is no difference in ammeter readings.