Question

(a) In Fig. 27-56, what current does the ammeter read if$\mathbf{\epsilon }\mathbf{=}\mathbf{5}\mathbf{.0}\mathbf{}\mathbf{\text{V}}$(ideal battery), ${\mathbf{R}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}Ω}}$, ${\mathit{R}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4.0}\mathbf{\text{\hspace{0.17em}Ω}}$, and ${\mathit{R}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6.0}\mathbf{\text{\hspace{0.17em}Ω}}$? (b) The ammeter and battery are now interchanged. Show that the ammeter reading is unchanged.

Short answer

1. The current shown in the ammeter reading is $0.45\text{\hspace{0.17em}A}$.
2. The ammeter reading is same as $0.45\text{\hspace{0.17em}A}$even if the ammeter and battery are now changed.

Step-by-step solution

The given data

Voltage reading across the ideal battery,$\epsilon =5.0\text{\hspace{0.17em}V}$

The value of the resistances given,${\text{R}}_1=2.0\text{\hspace{0.17em}Ω}$ , ${\text{R}}_2=4.0\text{\hspace{0.17em}Ω}$,and${\text{R}}_3=6.0\text{\hspace{0.17em}Ω}$

Understanding the concept of voltage, current and resistance

If we consider a loop then the current flowing through a given resistor can be given by the ratio of the total emf of the battery dividing by the equivalent resistance. Now, using this current value, we can get the current through the resistance on the same arm of the ammeter, and thus, it will give us the required ammeter reading value of the circuit.

Formulae:

The voltage equation using Ohm’s law, $\text{V}=\text{IR}$ (1)

The equivalent resistance for series combination of the resistors, ${\text{R}}_{\text{eq}}=\sum _{\text{i}}^{\text{n}}{\text{R}}_{\text{i}}$ (2)

The equivalent resistance for parallel combination of the resistors, ${\text{R}}_{\text{eq}}=\sum _{\text{i}}^{\text{n}}\frac{\text{1}}{{\text{R}}_{\text{i}}}$ (3)

a) Calculation of the ammeter reading

The current flowing through resistor can be given using equations (2) and (3) in equation (1) as follows:

$\begin{array}{c}{\mathrm{I}}_1=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_1+{\mathrm{R}}_2{\mathrm{R}}_3/({\mathrm{R}}_2+{\mathrm{R}}_3)}\\ =\frac{5.0\text{\hspace{0.17em}V}}{2.0\text{\hspace{0.17em}Ω}+\left(4.0\text{\hspace{0.17em}Ω}\right)\left(6.0\text{\hspace{0.17em}Ω}\right)/(4.0\text{\hspace{0.17em}Ω}+6.0\text{\hspace{0.17em}Ω})}\\ =\frac{5}{2+2.4}\text{A}\\ =1.14\text{\hspace{0.17em}A}\end{array}$

Now, for the ammeter reading, the current across resistor$R_3$ can be given using the above value in equation (1) as follows:

$\begin{array}{c}{I}_3=\frac{\epsilon -V_1}{R_3}\\ =\frac{\epsilon -I_1{R}_1}{R_3}\\ =\frac{5.0\text{\hspace{0.17em}V}-\left(1.14\text{\hspace{0.17em}A}\right)\left(2.0\text{\hspace{0.17em}Ω}\right)}{6.0\text{\hspace{0.17em}Ω}}\\ =0.45\text{\hspace{0.17em}A}\end{array}$

Hence, the ammeter reading is $0.45\text{\hspace{0.17em}A}$.

b) Calculation of the ammeter reading after interchanging the ammeter and battery

Now, the current flowing through resistor$R_3$ can be given using equations (2) and (3) in equation (1) as follows:

$\begin{array}{c}{I}_3=\frac{\epsilon }{R_3+R_1{R}_2/(R_1+R_2)}\\ =\frac{5.0\text{\hspace{0.17em}V}}{6.0\text{\hspace{0.17em}Ω}+\left(2.0\text{\hspace{0.17em}Ω}\right)\left(4.0\text{\hspace{0.17em}Ω}\right)/(2.0\text{\hspace{0.17em}Ω}+4.0\text{\hspace{0.17em}Ω})}\\ =\frac{5}{6+1.33}\text{A}\\ =0.682\text{\hspace{0.17em}A}\end{array}$

Now, for the ammeter reading, the current across resistor$R_1$ can be given using the above value in equation (1) as follows:

$\begin{array}{c}{I}_1=\frac{\epsilon -V_3}{R_1}\\ =\frac{\epsilon -I_3{R}_3}{R_1}\\ =\frac{5.0\text{\hspace{0.17em}V}-\left(0.682\text{\hspace{0.17em}A}\right)\left(6.0\text{\hspace{0.17em}Ω}\right)}{2.0\text{\hspace{0.17em}Ω}}\\ =0.45\text{\hspace{0.17em}A}\end{array}$

Hence, the ammeter reading is$0.45\text{\hspace{0.17em}A}$ and it is same as that of the above ammeter reading before the interchange.