wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In the circuit shown in figure E1=2E2=20V,R1=R2=10kΩ and C=1μF. The sum of current through R1,R2 and C when S has been kept connected to A for a long time.
216412_ab4c0eeb9b71429c9406b8e750997aa8.png

Open in App
Solution

When S is closed for long time, the capacitor C is fully charged and no current will pass through it. Thus, IC=0
The current drawn from battery E1 is I=E1R1+R2=20(10+10)103=1mA
As the resistances are in series so the current through them will be equal to I
Thus, IR1=IR2=1mA

529233_216412_ans_22d14a5c420043738ab0d4a07d63c229.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Energy of a Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon