Electric Bulbs
Trending Questions
Q. Five bulbs with power rating B1=B4=60 W−200 V, B2=B3=100 W−200V and B5=40 W−200V now, 200 V are connected as shown in figure. Then
1. B5 will glow the most
2. B1 will glow more than B4
3. B2 and B3 will glow the most
1. B5 will glow the most
2. B1 will glow more than B4
3. B2 and B3 will glow the most
- None
- 1, 2, and 3 are correct
- 1 and 2 are correct
- Only 1 is correct
Q. The balancing lengths of potentiometer wire are l1 and l2 when two cells of emf E1 and E2 are connected in the secondary circuit in series. first to support each other and next to oppose each other. Then the ratio of E1E2 is equal to [if E1>E2]
- l1−l2l1+l2
- l1+l2l1−l2
- l1l2
- l2l1
Q. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R is connected across the given cell has value of (i) infinity, (ii)9.5Ω the balancing length on the potentiometer wire are found to be 3 m and 2.85 m respectively. The value of internal resistance of the cell is
- 0.25 Ω
- 0.95 Ω
- 0.5 Ω
- 0.75 Ω
Q. A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The emf of the cell needed to make the bulb glow at full intensity is
- 4.5 V
- 1.5 V
- 2.67 V
- 13.5 V
Q. The circuit below is made up using identical light bulbs. The light bulb of maximum brightness among the following will be
- A
- C
- D
- E
Q. In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, E1E2 is:
- 53
- 85
- 32
- 43
Q. Which of the following acts as a circuit protection device?
- Conductor
- Inductor
- Switch
- Fuse
Q. In the given potentiometer circuit length of the wire AB is 3 m and resistance is R=4.5 Ω. The length AC for no deflection in galvanometer is
- 1.5 m
- 2 m
- 1.8 m
- 2.5 m
Q. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentionmeter of length 1 m and resistance 5 Ω. The value of R, to give a potential difference of 5 mV across 10 m of potentiometer wire is:
- 395 Ω.
- 495 Ω.
- 480 Ω.
- 490 Ω.
Q. Three bulbs B1, B2 and B3 are connected to the mains as shown in the figure. How will the brightness of bulb B1 be affected if B2 or B3 are disconnected from the circuit?
- Bulb B1 becomes brighter
- Bulb B1 becomes dimmer
- No change occurs in the brightness
- Bulb B1 becomes brighter if bulb B2 is disconnected and dimmer if bulb B3 is disconnected
Q. In the circuit diagram, all the bulbs are identical. Bulb will be the brightest.
- A
- B
- C
- D
Q. A potentiometer wire gives a null point at 3 m for a wire of length 3.5 m. The primary circuit contains only a battery of emf 10 V. The emf of secondary cell is 8 V. What is the internal resistance of the battery in the primary circuit (in ohms) if the resistance of the potentiometer wire is 14 Ω?
Q. In the circuit shown, a four - wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is x=0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be
- 0.20 V
- 0.75 V
- 0.25 V
- 0.50 V
Q. The length of a potentiometer wire 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of 20 Ω, the null point on it is found to be 1000 cm. The resistance of whole wire is:
- 80 Ω
- 120 Ω
- 60 Ω
- 100 Ω
Q. In a potentiometer wire experiment, the emf of a battery in the primary circuit is 20 Volt and its internal resistance is 5 Ω. There is a resistance box (in series with the battery and the potentiometer wire) whose resistance can be varied from 120 Ω to 170 Ω. Resistance of the potentiometer wire is 75 Ω. The following potential difference can not be measured using this potentiometer
- 5 V
- 6 V
- 7 V
- 8 V
Q. A potentiometer wire has length 10 m and resistance 10 Ω. It is connected to a battery of emf 11 V and internal resistance 1 Ω, then the potential gradient in the wire is
- 10 V/m
- 1 V/m
- 0.1 V/m
- none
Q. The voltage across a bulb is decreased by 2%. Assuming that the resistance of the filament remains unchanged the power of the bulb will
- decrease by 4%
- decrease by 2%
- increase by 4%
- increase by 2%
Q. A potentiometer wire has a length of 10 m and resistance 4 Ωm−1. An accumulator of emf 2 V and a resistance box are connected in series with it. Calculate the resistance to be introduced in the box to get a potential gradient of 0.1 V/m.
- 20 Ω
- 30 Ω
- 40 Ω
- 50 Ω
Q. A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. W1, W2and W3are the output powers of the bulbs B1, B2 and B3 respectively then
- W1 > W2 = W3
- W1 > W2 > W3
- W1 < W2 = W3
- W1 < W2 < W3
Q. Figure below shows a circuit which may be used to compare the resistance R of an unknown resistor with a 100 Ω standard. The distances l from one end of the potentiometer slider wire to the balance point are 400 mm and 588 mm when X is connected to Y and Z respectively. The length of the slide is 1.00 m. What is the value of resistance R?
- 32 Ω
- 147 Ω
- 47 Ω
- 68 Ω
Q. A potentiometer wire of length 10 m and resistance 30 ohm is connected in series with a battery of emf 2.5 V, internal resistance 5 ohm and external resistance R. If the fall of potential along the potentiometer wire is 50 mV/m the value of R in Ω is
Q. In the circuit diagram, all the bulbs are identical. Which bulb will be the brightest?
- A
- B
- C
- D
Q.
How does the strength of the magnetic field produced by a current-carrying circular coil depend on the number of turns in the coil?
Q. A potentiometer wire, 4 m long and resistance 12 Ω , is connected in series with a resistance of 1988 Ω and a cell of emf 2 V. What is the potential gradient along the wire ?
- 0.002 Vm
- 0.004 Vm
- 0.01 Vm
- 0.003 Vm
Q. A battery is connected to a potentiometer and the balance point is obtained at 84 cm along the wire. When its terminals are connected by a 5 Ω resistor, the balance point changes to 70 cm.
(i) Calculate the internal resistance of the cell.
[1 Mark]
(i) Calculate the internal resistance of the cell.
[1 Mark]
- 4 Ω
- 2 Ω
- 5 Ω
- 1 Ω
Q. n identical light bulbs, each designed to draw P power from a certain voltage supply, are joined in series across that supply. The total power which they will draw is
- nP
- P
- P/n
- P/n2
Q. Three bulbs B1, B2 and B3 are connected to the mains as shown in the figure. How will the brightness of bulb B1 be affected if B2 or B3 are disconnected from the circuit?
- Bulb B1 becomes brighter
- Bulb B1 becomes dimmer
- No change occurs in the brightness
- Bulb B1 becomes brighter if bulb B2 is disconnected and dimmer if bulb B3 is disconnected
Q. a circuit containing inductor, bulb and battery has a switch, the bulb will become suddenly bright if
- Contact is made or broken
- Contact is made
- Contact is broken
- Won't become bright at all
Q. A potentiometer wire of 4 m long having resistance 12 Ω is connected in series with a resistance of 1988 Ω and a cell of emf 2 V. What is the potential gradient along the wire ?
- 0.002 V/m
- 0.004 V/m
- 0.001 V/m
- 0.003 V/m
Q. AB is a potentiometer wire of length 100 cm and its resistance is 10 Ω. It is connected in series with a battery of emf 2 V and resistance 40 Ω. If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, the value of E is
- 0.8 V
- 1.6 V
- 0.08 V
- 0.16 V