TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Initially we have following circuit, when switch S is openTotal resistance of circuit=6+6=12Ω
Current in circuit,
I=‌

12V

12Ω

=1A
So, potential across R1 is
VAB=IAB×R1=1×6=6V
Now, after switch is closed we have following circuit,Total resistance is now,
Current through cell, I=‌

V

Req

=‌

12

9

=‌

4

3

A
Since, the value of resistance R1 and R2 are same as 6Ω, hence current I is equally divided through R1 and R2,
∴‌‌I1=I2=

I

2

=‌

4

3×2

=‌

2

3

A
So, potential across Rl is
VAB′=I1R1=‌

2

3

×6=4V
So, change in potential of R1 is
=VAB′−VAB=4−6=−2V