Question

assume the ammeter shown in figure 9 - 11 has an internal resistance of 60 ω with a maximum (full - scale) current of 25 ma. a shunt (parallel) resistance is required that will allow the meter to read a total of 250 ma by allowing 25 ma through the meter and the rest through the shunt. (a) what is the current in the shunt resistor? (b) specify the resistance and wattage rating of the shunt resistor.

Explanation:

Step1: Calculate current in shunt resistor

In a parallel - circuit for ammeter shunt, the total current $I_T$ is the sum of the current through the ammeter $I_M$ and the current through the shunt resistor $I_{sh}$. Given $I_T = 250\ mA$ and $I_M=25\ mA$, we use the formula $I_{sh}=I_T - I_M$.
$I_{sh}=250\ mA - 25\ mA=225\ mA$

Step2: Calculate shunt resistance

Since the voltage across the ammeter $V_M$ and the shunt resistor $V_{sh}$ is the same in parallel. First, find the voltage across the ammeter using Ohm's law $V = IR$. For the ammeter, $V_M=I_MR_M$, where $I_M = 25\times10^{- 3}\ A$ and $R_M = 60\ \Omega$. So $V_M=(25\times10^{-3}\ A)\times60\ \Omega = 1.5\ V$. Then, for the shunt resistor, using Ohm's law $R_{sh}=\frac{V_{sh}}{I_{sh}}$, and since $V_{sh}=V_M = 1.5\ V$ and $I_{sh}=225\times10^{-3}\ A$, we have $R_{sh}=\frac{1.5\ V}{225\times10^{-3}\ A}=\frac{1.5}{0.225}\ \Omega=\frac{20}{3}\Omega\approx6.67\ \Omega$.

Step3: Calculate wattage rating of shunt resistor

Use the power formula $P = VI$. For the shunt resistor, $P_{sh}=V_{sh}I_{sh}$. Substituting $V_{sh}=1.5\ V$ and $I_{sh}=225\times10^{-3}\ A$, we get $P_{sh}=1.5\ V\times225\times10^{-3}\ A = 0.3375\ W$.

Answer:

(a) The current in the shunt resistor is $225\ mA$.
(b) The resistance of the shunt resistor is $\frac{20}{3}\Omega\approx6.67\ \Omega$ and the wattage rating is $0.3375\ W$.