Question

An ammeter of resistance $0.6\Omega$ can measure current up to $1.0\;A$. Calculate
(i)The shunt resistance required to enable the ammeters to measure current up to $5.0\;A$
(ii)The combined resistance of the ammeter and the shunt.

Answer 1

For the first part, remember that a shunt resistance offers an alternative path to current flow owing to its low resistance. In other words, a shunt resistance is always connected in parallel with the circuit. Also, remember that voltage across the branches of components connected in parallel is the same. Using this information, calculate the combined resistance. As for the second part, use the expression you get in the first part for the net ammeter resistance and substitute the value of the shunt resistance that you find.

Formula used:
Ohm’s law for voltage across a resistor $V =IR$, where I is the current flowing through the resistor and R is the resistance.
The net resistance for two resistors $R_1$ and $R_2$ connected in parallel: $R_{net} = \dfrac{R_1 R_2}{R_1 +R_2}$

Complete answer: or Complete step by step answer:
We begin with an understanding that ammeter is an instrument that is used to measure currents in a circuit. Different ammeters have different ranges of the maximum and the minimum current they can measure.
Let us now proceed in the context of the question by listing out what has been given, and what we are required to find:
We have the internal resistance of the ammeter $R_A = 0.6\Omega$
The maximum current it can measure is $I_A=1.0A$
Therefore, the voltage across the ammeter is $V_A = I_A R_A = 1 \times 0.6 = 0.6V$
Now, it is required for the ammeter to measure currents up to $I = 5.0A$
This can be done by introducing a shunt resistance of resistance $R_{shunt}\Omega$
Therefore, the net resistance of the ammeter will be:

$\dfrac{1}{R_{net}} = \dfrac{1}{R_A} +\dfrac{1}{R_{shunt}} \Rightarrow R_{net} = \dfrac{R_A R_{shunt}}{R_A + R_{shunt}} = \dfrac{0.6R_{shunt}}{0.6 +R_{shunt}}$

Now, since they are connected in parallel the voltage across them remains the same:

$0.6 = I \times R_{net} = 5 \times \left(\dfrac{0.6R_{shunt}}{0.6 +R_{shunt}}\right) \Rightarrow 0.6(0.6 + R_{shunt}) = 0.6R_{shunt} \times 5 \Rightarrow 0.36 +0.6R_{shunt} = 3R_{shunt}$

$\Rightarrow (3-0.6)R_{shunt} = 0.36 \Rightarrow R_{shunt} = \dfrac{0.36}{2.4} = 0.15\Omega$
Therefore, the shunt resistance required to enable the ammeter measure currents upto $5.0A$ is $0.15\Omega$.

From the previous part, we have that the net resistance, which is the combined resistance of the ammeter with the shunt, is given by the expression:

$R_{net} = \dfrac{R_A R_{shunt}}{R_A + R_{shunt}} = \dfrac{0.6R_{shunt}}{0.6 +R_{shunt}}$
We have found that $R_{shunt} =0.15\Omega$
$\Rightarrow R_{net} =\dfrac{0.6 \times 0.15}{0.6 +0.15} = \dfrac{0.09}{0.75} = 0.12\Omega$
Therefore, we see that by introducing the shunt resistance, the combined resistance drops much lower than the initial ammeter resistance which was $0.60\Omega$.

Additional Information:
For an ammeter to effectively measure the current passing through the circuit, the drop in voltage has to be as little as possible, which means that it should be able to have the current pass through it with a minimum or a low resistance.
A shunt resistance is a resistive component of a circuit that creates an overall relatively low-resistance path for electric current to pass through. A shunt is thus, a path that the current alternatively takes owing to its low resistance. This is why we connect shunt resistance in parallel with the circuit.

Note:
As you may have noticed that in all circuit diagrams where an ammeter is employed to measure current, the ammeter is always connected in series in the circuit. This is necessary as the current flowing through all the circuit components and hence the ammeter is the same throughout when connected in series.
Also remember that, since an ammeter has a low internal resistance, it is not advisable to connect it in parallel with the circuit. This is because we know that current always follows a low resistance path, and this may lead to a maximum current passing through the ammeter which may burn the fuse or damage the ammeter.