Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.
(a) Will the bulbs in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
(a) Comparison of brightness in series and parallel:
In the series combination, the total resistance is three times that of one bulb. Hence the current in the series circuit is one-third of the current in the parallel circuit. Since power is proportional to \( I^2 R \), each bulb in the parallel combination receives higher current and therefore glows more brightly. Thus, the bulbs in the two circuits do not glow with the same brightness; the bulbs in parallel glow more brightly.
(b) Effect of one bulb getting fused:
In the series circuit, when one bulb gets fused, the circuit becomes open, current becomes zero, and none of the bulbs glow.
In the parallel circuit, each bulb has an independent path. When one bulb gets fused, the remaining bulbs continue to glow with the same brightness because their individual circuits remain complete.
State Ohm’s law. How can it be verified experimentally? Does it hold good under all conditions? Comment.
Statement of Ohm’s Law: Ohm’s law states that the potential difference \( V \) across a conductor is directly proportional to the current \( I \) flowing through it, provided physical conditions such as temperature remain constant. Thus, \( V \propto I \), or \( V = IR \).
Experimental Verification:
Validity of Ohm’s Law: Ohm’s law does not hold under all conditions. It fails when temperature changes significantly, in semiconductors, in filament lamps where resistance changes with heat, and in electrolytes where current flow is ionic. It applies only to ohmic conductors at constant temperature.
What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.
Electrical Resistivity: Electrical resistivity \( \rho \) of a material is defined as the resistance of a conductor of unit length and unit cross-sectional area. It is a measure of how strongly a material opposes the flow of electric current.
Unit: The SI unit of resistivity is ohm-metre (\( \Omega m \)).
Experiment to Study Factors Affecting Resistance:
The resistance \( R \) of a conductor depends on:
Using a circuit with a cell, ammeter, voltmeter, key and wires of different lengths, thickness and materials, readings of \( V \) and \( I \) are taken for each configuration. The resistance is calculated using \( R = V/I \). It is observed that resistance increases with length, decreases with increase in thickness, and varies for different materials even with identical dimensions.
How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?
Verification of equal current in a series circuit:
This confirms that the same current flows through every component in a series circuit because there is only one path for current flow.
How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?
Verification of equal potential difference in a parallel circuit:
This shows that in a parallel circuit, each resistor gets the same potential difference because all are connected directly across the same two points of the battery.
What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.
Joule’s Heating Effect: When an electric current \( I \) flows through a resistor \( R \) for time \( t \), the heat produced is given by \( H = I^2 R t \). This phenomenon is known as the Joule’s heating effect.
Demonstration: A circuit is set up with a battery, key, ammeter, and a coil immersed in water. When current flows through the coil, the temperature of water rises, demonstrating heat production due to current.
Applications:
Find the following in the electric circuit given:
(a) Effective resistance of two 8 Ω resistors in the combination
(b) Current flowing through 4 Ω resistor
(c) Potential difference across 4 Ω resistance
(d) Power dissipated in 4 Ω resistor
(e) Difference in ammeter readings, if any
(a) Effective resistance: The two 8 Ω resistors are in parallel. Their effective resistance is \( R = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{8 \times 8}{8 + 8} = 4\, \Omega \).
(b) Current through 4 Ω resistor: Total resistance in the circuit is \( 4 + 4 = 8\, \Omega \). Current \( I = \dfrac{V}{R} = \dfrac{8}{8} = 1\, \text{A} \).
(c) Potential difference across 4 Ω resistor: \( V = IR = 1 \times 4 = 4 \text{ V} \).
(d) Power dissipated: \( P = I^2 R = 1^2 \times 4 = 4 \text{ W} \).
(e) Difference in ammeter readings: There is no difference because the same current flows through all elements of a series circuit.