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Sagot :
To solve for the value of the shunt resistor [tex]\( R_s \)[/tex] required to convert a galvanometer into an ammeter, we need to follow a series of calculations based on the information provided.
Given:
- Full scale current of the galvanometer ([tex]\( I_g \)[/tex]) = 12.0 mA = 0.012 A
- Coil resistance ([tex]\( R_c \)[/tex]) = 60.0 Ω
- Maximum current to be measured by the ammeter ([tex]\( I_{max} \)[/tex]) = 0.500 A
We need to find:
- The value of the shunt resistor ([tex]\( R_s \)[/tex]).
Step-by-step solution:
1. Understand the relationship:
The shunt resistor is used to divert most of the current away from the galvanometer because the galvanometer can only handle a small current. The total current [tex]\( I_{max} \)[/tex] will split into two parts: one part through the galvanometer ([tex]\( I_g \)[/tex]) and the other through the shunt resistor ([tex]\( I_s \)[/tex]).
2. Calculate the current through the shunt resistor ([tex]\( I_s \)[/tex]):
Since the galvanometer can handle only [tex]\( I_g \)[/tex]:
[tex]\[ I_s = I_{max} - I_g = 0.500 \, \text{A} - 0.012 \, \text{A} = 0.488 \, \text{A} \][/tex]
3. Apply Ohm's Law to find [tex]\( R_s \)[/tex]:
The voltage drop across the shunt resistor ([tex]\( R_s \)[/tex]) must be the same as the voltage drop across the coil resistance ([tex]\( R_c \)[/tex]) of the galvanometer, because they are in parallel.
The voltage drop across the galvanometer is:
[tex]\[ V_g = I_g R_c = 0.012 \, \text{A} \times 60.0 \, \Omega = 0.72 \, \text{V} \][/tex]
Therefore, the same voltage drop must occur across [tex]\( R_s \)[/tex]:
[tex]\[ V_s = I_s R_s \][/tex]
Since [tex]\( V_g = V_s \)[/tex]:
[tex]\[ 0.72 \, \text{V} = 0.488 \, \text{A} \times R_s \][/tex]
4. Solve for [tex]\( R_s \)[/tex]:
[tex]\[ R_s = \frac{0.72 \, \text{V}}{0.488 \, \text{A}} \approx 1.475 \, \Omega \][/tex]
5. Compare with given choices:
The closest available choice to our calculated [tex]\( R_s \)[/tex] is [tex]\( 1.44 \, \Omega \)[/tex] (choice [tex]\( (d) \)[/tex]).
Conclusion:
- The value of the shunt resistor [tex]\( R_s \)[/tex] should be approximately [tex]\( 1.475 \, \Omega \)[/tex], and the closest given option is:
[tex]\[ \boxed{1.44 \, \Omega} \][/tex]
Given:
- Full scale current of the galvanometer ([tex]\( I_g \)[/tex]) = 12.0 mA = 0.012 A
- Coil resistance ([tex]\( R_c \)[/tex]) = 60.0 Ω
- Maximum current to be measured by the ammeter ([tex]\( I_{max} \)[/tex]) = 0.500 A
We need to find:
- The value of the shunt resistor ([tex]\( R_s \)[/tex]).
Step-by-step solution:
1. Understand the relationship:
The shunt resistor is used to divert most of the current away from the galvanometer because the galvanometer can only handle a small current. The total current [tex]\( I_{max} \)[/tex] will split into two parts: one part through the galvanometer ([tex]\( I_g \)[/tex]) and the other through the shunt resistor ([tex]\( I_s \)[/tex]).
2. Calculate the current through the shunt resistor ([tex]\( I_s \)[/tex]):
Since the galvanometer can handle only [tex]\( I_g \)[/tex]:
[tex]\[ I_s = I_{max} - I_g = 0.500 \, \text{A} - 0.012 \, \text{A} = 0.488 \, \text{A} \][/tex]
3. Apply Ohm's Law to find [tex]\( R_s \)[/tex]:
The voltage drop across the shunt resistor ([tex]\( R_s \)[/tex]) must be the same as the voltage drop across the coil resistance ([tex]\( R_c \)[/tex]) of the galvanometer, because they are in parallel.
The voltage drop across the galvanometer is:
[tex]\[ V_g = I_g R_c = 0.012 \, \text{A} \times 60.0 \, \Omega = 0.72 \, \text{V} \][/tex]
Therefore, the same voltage drop must occur across [tex]\( R_s \)[/tex]:
[tex]\[ V_s = I_s R_s \][/tex]
Since [tex]\( V_g = V_s \)[/tex]:
[tex]\[ 0.72 \, \text{V} = 0.488 \, \text{A} \times R_s \][/tex]
4. Solve for [tex]\( R_s \)[/tex]:
[tex]\[ R_s = \frac{0.72 \, \text{V}}{0.488 \, \text{A}} \approx 1.475 \, \Omega \][/tex]
5. Compare with given choices:
The closest available choice to our calculated [tex]\( R_s \)[/tex] is [tex]\( 1.44 \, \Omega \)[/tex] (choice [tex]\( (d) \)[/tex]).
Conclusion:
- The value of the shunt resistor [tex]\( R_s \)[/tex] should be approximately [tex]\( 1.475 \, \Omega \)[/tex], and the closest given option is:
[tex]\[ \boxed{1.44 \, \Omega} \][/tex]
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