Ques 13. A 200 W, 200 V bulb and a 100 W, 200 V bulb are connected in series and the voltage of 400 V is applied across the series-connected bulbs, Under this condition
100 W bulb will be brighter than 200 W bulb✔
200 W bulb will be brighter than 100 W bulb
Both the bulbs will have equal brightness
Both the bulbs will be darker than when they are connected across rated voltage
To check the brightness of the lamp first we need to calculate the resistance of each lamp
For 200W, 200V bulb 1 the resistance is
P = V2/R or R = V2/P
R = (200)2/200 = 200Ω
For 100 W, 200 V bulb 2 the resistance is
R = (200)2/100 = 400Ω
Since the bulb are connected in series the total resistance will be
Req = 400 + 200 = 600Ω
I = V/Req = 400/600 = 2/3A
Since the bulbs are in series, the same amount of current flows through each of them.
Now note that the potential across each bulb, in this case, is not 400 volts. Instead, it is the voltage across the combination of bulbs.
The power at which each bulb is working is given by
P = V × I
If we substitute V = IR from Ohm’s law, we get
P = I2R
For bulb 1 power is
P = 200 × (2/3)2
= 800/9 W
For bulb 2 power is
P = 400 × (2/3)2
= 1600/9 W
Since the power of bulb 2 i.e 100 W bulb is more than the power of bulb1 200 W bulb. Hence the 100 W bulb will be brighter than the 200 W bulb.
Note:- Also It can be seen that the resistance of the 100W bulb is greater than the resistance of the 200 W bulb. Hence, the heat produced in the 100W bulb is more than produced in 200 W.Conclusion: 100W will glow brighter than the 200W bulb.
Ques 14. In the network shown, if one of the 4Ω resistances is disconnected, when the circuit is active, the current flowing now will
Increased Very much
Decreased✔
Zero
Increase very slightly
If we do not remove the 4Ω resistance then the equivalent resistance is 10Ω
But if we remove one of the 4Ω resistance then the equivalent resistance becomes = 12Ω
Hence if any of the 4Ω resistance is disconnected then the total resistance is increased so the current decreases.
Ques 15. For the circuit shown in the figure, when Vs = 0, I = 3A, When Vs = 200 V, what will be the value of I?
-4A
-1A✔
1A
7A
Applying the current division rule when Is is short-circuited
Ques 16. For the linear circuit shown in the figure,
when R = ∞, V = 20 V;
when R = 0, I = 4 A;
when R = 5 Ω the current I is
1 A
2 A✔
3 A
4 A
R = V/I
R’ = 20/4 = 5Ω
When R = 5Ω then current I is
I = V/R + R’
I = 20/5+5 = 2A
Ques 17. The current I in the circuit shown in the figure is
-3.67 A
-1 A✔
4 A
6 A
Let the current passes from 2Ω resistance is i then
2i + 2(6 + i) – 8(3-i) = 0
4i + 12 +2i – 24 + 8i = 0
12i = 12
i = 1
Hence I = -i = -1A
Ques 18. In the network shown in the figure, the value of R1 such that maximum possible power will be transferred to RL is
5.76Ω
5.97 Ω✔
10.0 Ω
15.0 Ω
Disconnect the load resistance from the load terminals A and B
Thevenin’s equivalent resistance or resistance across the terminals AB is
From the maximum power transfer theorem, the RL value must equal the RTH to deliver the maximum power to the load.
Therefore, RL = RTH= 5.97 Ohm
Ques 19. A resistance R is measured by the ammeter-voltmeter method. The voltmeter reading is 200 V and its internal resistance is 2 K. If the ammeter reading is found to be 2 A, then the value of R is
105.3 Ω✔
100.0 Ω
95.3 Ω
90.3 Ω
The Value of voltmeter across Rx is given as
Ques 20. The circuit shown in the figure is equivalent to the load of
