Class 10 Science
Question 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.
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:
This can be expressed by following equation:
`H=I^2` RtWhere, 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.
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
(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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