Question

Look at the circuit shown in the figure above. with switch s1 closed and the resistance value of r1 and r2 equal, meter m1 has a current reading of 0.04 A. what will be the current readings on meter M2 and meter M3? A. Meter M2 will read 0.02 A, and meter M3 will read 0.02 A.

B. Meter M2 and meter M3 will both read 0.1 A.
C. Meter M2 and meter M3 will both read 0.04 A.
D. Meter M2 will read 0.2 A, and meter M3 will read 0.4 A.​

Answer 1

With switch S1 closed and equal resistance values of R1 and R2, the current reading on meter M2 and meter M3 will be 0.02 A each.

In the given circuit, with switch S1 closed and the resistance values of R1 and R2 equal, meter M1 has a current reading of 0.04 A.

Since R1 and R2 have equal resistance values and are in parallel, they will share the total current flowing through them equally.

This means that both R1 and R2 will have a current reading of 0.02 A.

Therefore, the current readings on M2 and M3 will also be 0.02 A each.

Therefore, the correct option is A. Meter M2 will read 0.02 A, and meter M3 will read 0.02 A.

Answer 2

A. Meter M2 will read 0.02 A, and meter M3 will read 0.02 A

Step-by-step explanation:

I = Current through M1,

I1 = Current through M2

I2 = Current through M3

According to current division rule,

I1 = { R1 / ( R1 + R2) } * I ------------------------------( 1 )

And similarly

I2 = { R2 / ( R1 + R2) } * I -------------------------------( 2 )

and hence when the Switch S1 is closed and given that the Resistance Value of R1 and R2 becomes equal then the Equation ( 1 ) and equation ( 2 )

becomes :

==> I1 = { R1 / ( R1 + R1) } * I

==> I1 = { R1 / 2( R1 ) } * I

==> I1 = (0.5) * I

and as the value of I is given as 0.04A

therefore:

the value of I1 is 0.04 / 2 = 0.02 A ,

and similarly the value of I2 is also 0.02 A.