Question

A wire of resistance R connected in series with 1.5V cell is found to be carrying a current of 0.05A. If the wire is now connected i parallel with an identical wire, find the new current in the circuit

Answer 1

Answer:Assuming, The new circuit is like the circuit attached below, the current in the new circuit will be 0.1 Ampere.

Explanation: Given, We have Potential Difference of 1.5V, Current of 0.05A.

Now, we know V = I x R ,where V is potential difference, I is current and R is resistance in the circuit.

∴ Putting the values ⇒ 1.5V = 0.05A x R

⇒R = 1.5/0.05

∴R = 30 Ω

Now, It is said that A same identical wire is connected in parallel with the previous wire.

∴ New R will be ⇒
`(1)/(R(new))` =
`(1)/(r) + (1)/(r)`

=[1/30] + [1/30]

=1/15

∴ 1/R(new) = 15 Ω

Now, New Current in the circuit will be:

⇒ V= IR

⇒1.5V = I x 15

⇒I = 0.1 A

Therefore, Answer will be 0.1 ampere.