A resistor of resistance R is connected to a battery of negligible internal resistance. If a siilar resistor is connected in series with it the

a

effective resistance of the circuit is halved

b

total power dissipated is doubled

c

total current in the circuit is halved

d

terminal voltage is halved

Download Offline App Ask a Question

Explanation

Correct Option
b

No explanation available

Video Explanation

No video available

Post your Contribution

Share:

Discussions (13)

titan52
6 years ago

I=E/(R+r) r is negligible in the question

So...I=E/(R). Another similar resistor is added in series; I=E/(R+R)
I=E/2R . Therefore current is reduced by 2
Ans: C

Dogty
6 years ago

P=VI
P=I²R
P=I²(R+R)
P=I²×2R
P=2I²R
Power is doubled

Alexvictory2
2 years ago

When a resistor of resistance R is connected to a battery with negligible internal resistance and then a similar resistor is connected in series with it, the effective resistance of the circuit is increased to 2R. This is because resistors in series add up to give the total resistance.

Applying Ohm's Law (V = IR), since the resistance of the circuit has doubled, the total current in the circuit will be halved (C). This is because, with the same voltage applied by the battery, as resistance increases, current decreases.

The total power dissipated in the circuit depends on the voltage, current, and resistance. Since the current is halved but the total resistance has doubled, the total power dissipated will be halved, not doubled (B).

The terminal voltage across each resistor will also be halved in this scenario. This is because the total voltage supplied by the battery is distributed across both resistors, and adding another resistor in series effectively divides the voltage equally across them, resulting in a halving of the terminal voltage (D).

Ai

Domnickado
3 years ago

You guys saying 'power is doubted' are actually forgetting that the current is also affected. If your calculation is right the current and power are both halved, so the answer should be C

agubanze
1 year ago

ok

PRadiant
2 years ago

I'm thinking B should actually be correct, since it is a Series Connection.
Yuuno, being a series connection, the same current flows through the resistors, so, the current should not be halved or doubled.
It should be the same, but Power is affected.

Frankyjay
3 years ago

total resistor, R=R+R=2R
but V=IR=2IR
P=IV=2I²R
P=2(I²R)
total power is doubled

TheBigJo100
3 years ago

Actually The resistance is doubled and the current is halved so the power remains the same

Quick Questions

Ask a Question
CO

ceoofwahala

20th June, 2026

Chemistry


2 comments

ASSAAS

20th June, 2026

English Language


5 comments

infinitehoaxx

21st May, 2026

Computer


4 comments