Back to Directory
NEET PHYSICSMedium

10 resistors, each of resistance RR are connected in series to a battery of emf EE and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased nn times. The value of nn is

A

10

B

100

C

1

D

1000

Step-by-Step Solution

In series, total resistance Rs=10RR_s = 10R. Current Is=E/(10R)I_s = E / (10R). In parallel, total resistance Rp=R/10R_p = R/10. Current Ip=E/(R/10)=10E/RI_p = E / (R/10) = 10E/R. The ratio Ip/Is=(10E/R)/(E/10R)=100I_p / I_s = (10E/R) / (E/10R) = 100. Thus n=100n = 100.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started