Question

In the circuit shown in Fig. E26.23, ammeter A1 readsand the batteries have no appreciable internal resistance.

(a) What is the resistance of R? (b) Find the readings in the other ammeters.

Short answer

the resistance is 20 ohms

the readings are${\mathrm{I}}_2=4\mathrm{A},{\mathrm{I}}_3=12\mathrm{A},{\mathrm{I}}_4=14{\mathrm{A}}_5=8\mathrm{A}$

Step-by-step solution

About ammeters

Instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.

Determie the resistance 

(a)Ammeterl(A1)readscurrent.

${\mathrm{I}}_1=10\mathrm{A}$The resistance R is in series with the remaining of the circuit (right part of
circuit), so the current flow in R is due to the voltage drop where the voltage drop is equal to the emf of the battery as ‘r = 0
V = 200 V
We can get the value of R by using Ohm's law where V is the voltage of 220 V battery and current through R is the current
that measured by A1
$$\begin{gathered} \mathrm{V}=200\mathrm{V} \\ \mathrm{R}=\frac{\mathrm{V}}{\mathrm{l}} \\ =\frac{200}{10} \\ =20\mathrm{\Omega } \end{gathered}$$

Therefore the resistance is 20 ohms

Determine the readings in ammerter

(b)The?gurebelowshoWsthesituationwheretheredarrow showthedirectionsofthecurrentTheammeterAreadsthe
current on resistance 20 ohms where this current is due to the voltage drop in 160 V battery. So, we could use Ohm's law to get
the current I measured by A in the next form

$$\begin{gathered} {\mathrm{I}}_5=\frac{\mathrm{V}}{20} \\ =\frac{160}{20} \\ =8\mathrm{A} \\ {\mathrm{I}}_2=\frac{160}{40} \\ =4\mathrm{A} \end{gathered}$$

WiththesameconceptofA5wecangetthecurrentI2measuredbyA2wherethevoltagedropin160Vbatteryisdueto
the resistance 40 0

$$\begin{gathered} {\mathrm{I}}_3={\mathrm{I}}_2+{\mathrm{I}}_5 \\ =4+8 \\ {\mathrm{I}}_4={\mathrm{I}}_1+{\mathrm{I}}_2 \\ =10+4 \\ =14\mathrm{A} \end{gathered}$$

the current measured by A3 and A4, we will use the junction rule.
Again, but at point from point a so the current I4 is the sum
of both current

$$\begin{gathered} {\mathrm{I}}_3={\mathrm{I}}_2+{\mathrm{I}}_5 \\ =4+8 \\ {\mathrm{I}}_4={\mathrm{I}}_1+{\mathrm{I}}_2 \\ =10+4 \\ =14\mathrm{A} \end{gathered}$$

Therefore the readings are${\mathrm{I}}_2=4\mathrm{A},{\mathrm{I}}_3=12\mathrm{A},{\mathrm{I}}_4=14{\mathrm{A}}_5=8\mathrm{A},$