A cell of e.m.f 1.5V and internal resistance 2.552 is connected in series with an ammeter resistance 0.50 and a load of resistance 7.00. Calculate the current in the circuit

Pacifica

3 years ago

We are given resistance connected in series.Therefore;

R= R1+R2+R3
=2.552+0.50+7.0= 10.1

And we know that V=IR
V=1.5
R= 10.1
We are told to look for the current I.

1.5=10.1I
And
I=1.5/10.1=0.148~0.15.

12funmilola13

1 year ago

E.m.f = I (R+r)

E.m.f = 1.5v
r = 2.552
R = 0.50+7.00 = 7.5

I = ?

1.5 = I (7.5+2.552)
1.5 = I (10.0052)

I = 1.5/10.0052
I = 0.1499
= 0.15