A current of 0.6 a is shown by ammeter

Asked by admin @ in Physics viewed by 439 People

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.

Answered by admin @


When k₁ is closed : only Current passing through 5Ω resistor and lamp of resistance R ( Let). There is zero current passing through 10Ω.
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

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