Question

An unknown resistance X Is connected in parallel to 20ohms resistance .with 2volts and other resistance in series with current flowing .o5in x​

Answer 1

Answer:

Let R be the combined resistance of galvanometer and an unknown resistance in series and r be the internal resistance of each battery. When the batteries, each of emf E are connected in series, the net emf = 2E and net internal resistance = 2r.

Current i

1

=

R+2r

2E

or 1.0=

R+2r

2×1.5

or R+2r=3 (i)

When the batteries are connected in parallel, the emf remains E and net internal resistance become r/2. Therefore, current is:

i

2

=

R+

2

r

E

=

2R+r

2E

or 2R+r=

i

2

2E

=

0.6

2×1.5

=5.0 (ii)

Solving Equations (i) and (ii), we get r=

3

1

Ω

From question, Value of x=3

Answered By

Explanation:

Answer By WANIIQRA