Assertion (A): The rate at which energy is being delivered to a light bulb is lower after it has been on for a few seconds than just after it is turned on.
Reason (R): As the filaments warms up, its resistance rises and the current falls.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: Resistance of metals increases with temperature. Power \(P = V^2/R\).
Formula: \(P = V^2/R\), \(R \propto T\).
Solution: As filament warms, its resistance \(R\) increases. For constant voltage \(V\), current \(I = V/R\) decreases, so power \(P = V^2/R\) delivered to the bulb also decreases. Thus, A and R are true and R explains A.
Assertion (A): Two identical cells are connected in (a) series (b) parallel then maximum power transferred to the load is same in both cases.
Reason (R): Value of load resistance for maximum power transfer for series and parallel combination of cells are same.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: Maximum power transfer theorem for cells.
Formula: For a source with EMF \(E_{eq}\) and internal resistance \(r_{eq}\), maximum power is transferred when load \(R_L = r_{eq}\), and \(P_{max} = \frac{E_{eq}^2}{4r_{eq}}\).
Solution: For 2 cells in series: \(E_{eq} = 2E\), \(r_{eq} = 2r\). \(P_{max, series} = \frac{(2E)^2}{4(2r)} = \frac{E^2}{2r}\). Load \(R_L = 2r\).
For 2 cells in parallel: \(E_{eq} = E\), \(r_{eq} = r/2\). \(P_{max, parallel} = \frac{E^2}{4(r/2)} = \frac{E^2}{2r}\). Thus, maximum power transferred is same, so A is true. Load resistances for max power transfer are different, so R is false.
Assertion (A): The rate at which energy is being delivered to a light bulb is lower after it has been on for a few seconds than just after it is turned on.
Reason (R): As the filaments warms up, its resistance rises and the current falls.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Initially, the bulb's cold filament has lower resistance. As it heats up, resistance increases, causing current to drop (for a constant voltage source). Since power \(P = V^2/R\), an increase in \(R\) leads to a decrease in \(P\). Therefore, both assertion and reason are true, and the reason correctly explains the assertion.
Assertion (A): power consumed in circuit is maximum when current in circuit is maximum.
Reason (R): Current in circuit is maximum when power consumed by load is maximum.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Power consumed by the external resistor (R) is \(P = I^2R = \left(\frac{E}{R+r}\right)^2 R\). Maximum power is delivered to the load when the external resistance equals the internal resistance R=r, according to the maximum power transfer theorem. Maximum current occurs when (R=0). Therefore, both Assertion (A) and Reason (R) are false.
Assertion (A): In the given circuit, \(r\) is variable, value of \(I\) is maximum when \(r = R\).
Reason (R): At \(r = R\) power produced across \(R\) is minimum.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
The current in the circuit is \(I = E / (R + r)\). For \(I\) to be maximum, the denominator \((R + r)\) must be minimum. This occurs when \(r = 0\), not \(r = R\). So, Assertion (A) is false. The power produced across \(R\) is \(P_R = I^2 R = (E / (R + r))^2 R\). As \(r\) increases, \(R+r\) increases, so \(P_R\) decreases. Thus, \(P_R\) is minimum when \(r\) is maximum (approaching infinity), not at \(r = R\). So, Reason (R) is also false.
Assertion (A): The brightness of light bulb in a room decreases when heavy current appliance is switched on.
Reason (R): There will be no change in brightness of bulb if source is ideal and for non ideal source voltage drop across bulb decreases.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
When a heavy current appliance is switched on, it draws a large current from the power supply. For a non-ideal source with internal resistance, this increased total current causes a larger voltage drop across the internal resistance of the source (or wiring). Consequently, the terminal voltage supplied to other devices like the light bulb decreases. A lower voltage across the bulb reduces its power output (\(P = V^2/R\)) and thus its brightness. So, Assertion (A) is true. Reason (R) correctly states that brightness remains constant for an ideal source and that voltage across the bulb decreases for a non-ideal source, which explains the dimming. Thus, Reason (R) is true and explains Assertion (A).
Assertion (A): \(100 \text{ W}\), \(60 \text{ W}\) and \(20 \text{ W}\) bulbs, each marked \(220 \text{ volt}\), are connected in series with a voltage source, then \(20 text{ W}\) bulb gives maximum illumination.
Reason (R): Resistance of filament \(20 \text{ W}\) bulb is maximum.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
The power rating of a bulb is given by \(P = V^2 / R\). For bulbs rated at the same voltage \(V\) (here \(220 text{ V}\), resistance \(R = V^2 / P\). A lower power rating implies higher resistance. Thus, the \(20 \text{ W}\) bulb has the highest resistance (Reason R is true). When bulbs are connected in series, the same current \(I\) flows through each. The power dissipated by each bulb is \(P_{actual} = I^2 R\). Since \(I\) is common, the bulb with the highest resistance will dissipate the most power and therefore glow brightest. Hence, the \(20 \text{ W}\) bulb will provide maximum illumination (Assertion A is true). Reason (R) correctly explains Assertion (A).
Assertion (A): Though the same current flows through the line wires and the filament of the bulb but the rate of heat produced in the filament is much higher than that in line wires.
Reason (R): The filament of bulbs is made of a material of high resistance and low melting point.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true because heat produced is \( H = I^2 R t \) and the filament has a much higher resistance \( R \). Reason (R) is false because filaments (like tungsten) have a *high* melting point to withstand high temperatures.
Assertion (A): The coil of a heater is cut into two equal halves and only one of them is used into heater. The heater will now require half the time to produce the same amount of heat.
Reason (R): The heat produced is directly proportional to the square of current.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
When the heater coil is cut in half, its resistance becomes \( R' = R/2 \). Since \( H = \frac{V^2}{R}t \), for the same heat \( H \) and constant voltage \( V \), the time \( t \) is proportional to \( R \). Halving \( R \) halves \( t \). So (A) is true. (R) is also true by Joule's law, but it does not explain (A) because (A) implicitly assumes constant voltage, where \( I \) would change.
Assertion (A): In series combination of electrical bulb, lower power bulb emits more light than that of higher power bulb.
Reason (R): The lower power bulb in series gets more current than the higher power bulb.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
In a series circuit, the current \( I \) is the same for all components. The power dissipated by a bulb is given by \( P = I^2 R \). A lower power bulb has a higher resistance \( (R = V^2/P) \). Thus, it dissipates more power and glows brighter in series. Reason (R) is false because the current is the same for all bulbs in series.