Electricity Exemplar Problems Long Answer Questions Part 2 Class 10 Science

Electricity

Exemplar Problems

Long Answer Questions 2

33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Answer: Take three resistors R1, R2 and R3, a battery, an ammeter, a voltmeter and some wires. Complete the circuit; as shown in the diagram.

electricity exemplar problems question circuit diagram 12

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:

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.


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

electricity exemplar problems question circuit diagram 13

(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


(c) Potential difference across 4 Ω resistance

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.



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