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Question : 29
Total: 30
(a) For the combination of resistors shown in the following figure, find the equivalent resistance between M & N .5 A fuse for an electric iron which consumes 1 kW power at 220 V ?(d) Why is it impracticable to connect an electric bulb and an electric heater in series?
(b) State Joule's law of heating.(c) Why we need a
Solution:
(a) ( R 3 × R 4 ∕ R 3 + R 4 ) + R 1 + R 2
(b) Joule's law of heating implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current and (iii) directly proportional to the time for which the current flows through the resistor.
H = I 2 Rt
where
I = Current
R = Resistanct
T = Time taken
(c) For an electric iron which consumes1 kW electric power when operated at 220 V , a current of ( 1000 ∕ 220 ) A , i.e., 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.
(d) In a series circuit the current is constant throughout the electric circuit. Thus, it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly.
(b) Joule's law of heating implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current and (iii) directly proportional to the time for which the current flows through the resistor.
where
(c) For an electric iron which consumes
(d) In a series circuit the current is constant throughout the electric circuit. Thus, it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly.
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