A current of 0.6 A is shown by ammeter in the circuit when the K1 is closed.Find the resistance of lamp L. What change in the current flowing through the 5ohm and potential difference across the lamp will take place if the k2 is also closed .Give reason for your answer.
Answers
Also we can see 5Ω and RΩ are in series
So, Req = 5Ω + RΩ
Now, use Ohm's law,
V = IR
⇒ 6 = 0.6 × ( 5 + R) [ ∵ i = 0.6A is given ]
⇒ 10 = 5 + R
⇒ R = 5Ω
Hence, resistance of lamp is 5Ω
when K₂ is also closed :- Let i current flows through the circuit.
Req = 10Ω × 10Ω/(10Ω + 10Ω) = 5Ω [ because two resistors each of resistance 5Ω are joined in series and and both are joined with 10Ω resistor in parallel ]
Now, Req = 5Ω
so, current through the circuit is i = V/R = 6/5 = 1.4A
So, current through the resistor 5Ω is half of 1.4A [ ∵ 10Ω resistor and (5Ω + 5Ω) resistors are connected in parallel . And we know , potential in parallel combination is same . So, Current will be half due to each of resistors have same resistance e.g. 10Ω ]
so, current through 5Ω resistor is 0.7A
And current through lamp is also 0.7A , so potential difference across lamp is 0.7 × 5 = 3.5Volt
Explanation:
when k1 is closed and k2 is opened
I1=0.6A
R1=5ohm
RL(resistance of lamp)= ?
V=6v
according to ohms law,
V=IRs (here, Rs=R1+RL)
V=IRs
6=0.6*Rs
Rs=6/0.6
=10
R1+RL=10
5+RL=10
therefore,
RL=10-5
RL=5ohm
When K1 and K2 is closed
V=IR
potential difference across lamp =VL
VL=IRL
=0.6*5
=3v
When 10ohm resistor is connected in parallel with series combination of R1and RL (K1and K2 are closed) ,potential difference remains same across them.
So,
potential difference =I 2*R2=I 1*Rs
I1/I2 = R2/Rs
R2/Rs=R2/R1+RL
I1/I2=10/5+5
=10/10=1
therefore, I1/I2=1
=== I1 =I2
I/R=1/R2+1/Rs
=1/10+1/10
1/R =1/5
R= 5ohm
total current drawn from. the battery,
I=V/R
=6/5
=1.2A
we know that,
I=I1+I2
=2I1
1.2=2I1
I1=1.2/2=0.6A
so, there will be no change in the current and potential difference across the lamp.