The coil of wire in a galvanometer has a resistance of RC = 78.0 Ù. The galvanometer exhibits a full-scale deflection when the current through it is 0.440 mA. A resistor is connected in series with this combination so as to produce a voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of 21.0 V. What is the resistance of this resistor?
Vc = I*Rc = 0.44*10^-3 * 78=0.0343 Volts
R = Vr/I = (21-0.0343)/0.44*10^-3 = 47,649 Ohms.
R+78 = 21/0.44*10^-3 = 47,727.3,
R = 47,649.3 Ohms.
To find the resistance of the resistor connected in series with the galvanometer, we can use the concept of voltage division.
First, let's determine the current flowing through the galvanometer when it exhibits a full-scale deflection. Given that the resistance of the coil in the galvanometer is RC = 78.0 Ω and the full-scale current is 0.440 mA, we can use Ohm's Law:
Voltage (V) = Current (I) * Resistance (R)
Rearranging the equation to solve for current, we get:
I = V / R
Substituting the given values, we can calculate the full-scale current through the galvanometer:
I = 0.440 mA = 0.440 * 10^(-3) A
RC = 78.0 Ω
I = V / RC
0.440 * 10^(-3) A = V / 78.0 Ω
V = 78.0 Ω * 0.440 * 10^(-3) A
V = 0.03432 V
So, when the galvanometer exhibits a full-scale deflection, the voltage drop across it is 0.03432 V.
Now, let's calculate the resistance of the series resistor needed to produce a full-scale deflection of 21.0 V on the voltmeter.
To achieve voltage division, we can relate the current flowing through the galvanometer (0.440 mA) to the current flowing through the resistor (I_resistor). Since the two elements are in series, the total current is the sum of the currents through each component:
I_total = I_galvanometer + I_resistor
Since the full-scale deflection on the voltmeter is 21.0 V, the current flowing through the voltmeter (I_total) can be calculated using Ohm's Law:
I_total = V_total / (RC + R_resistor)
Now, let's solve for R_resistor:
R_resistor = (V_total / I_total) - RC
Substituting the given values:
V_total = 21.0 V
I_total = I_galvanometer = 0.440 * 10^(-3) A
RC = 78.0 Ω
R_resistor = (21.0 V / 0.440 * 10^(-3) A) - 78.0 Ω
R_resistor = 47818 Ω
Therefore, the resistance of the resistor connected in series to the galvanometer is approximately 47818 Ω.