Question

Choose the correct option/options:

Three ammeters A, B and C of resistances $R_A$, $R_B$ and $R_C$ respectively are joined as shown in the figure. When some potential difference is applied across the terminals $T_1$ and $T_2$, their readings are $I_A$, $I_B$ and $I_C$ respectively:

• A. $I_A=I_B$
• B. $I_A{R}_B+I_B{R}_B=I_C{R}_C$
• C. $\frac{I_A}{I_C}=\frac{R_C}{R_A}$
• D. $\frac{I_B}{I_C}=\frac{R_C}{R_A+R_B}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $I_A=I_B$
B $I_A{R}_B+I_B{R}_B=I_C{R}_C$
D $\frac{I_B}{I_C}=\frac{R_C}{R_A+R_B}$

Ammeters are connected in series and it gives the measurement of current passing through their path.

therefore, $I_A=I_B$...(1) because A and B are connected in series.

now, $V_{t2}-V_{t1}=I_A\ast R_A+I_B\ast R_b$...(2)

also $V_{t2}-V_{t1}=I_c\ast R_c$...(3)

by equating eq (2) and eq.(3) we get $I_A{R}_A+I_B{R}_B$. $=I_C{R}_C$....(4) now if in eq. (4) we put $I_A=I_B$ then,

$\therefore \frac{I_B}{I_C}=\frac{R_C}{R_A+R_B}$

hence option A , B and D are correct.