A 65 V. battery is connected across a 65400 ohm and a 37700 ohm resistor connected in series. What is the per cent error in the voltmeter reading when placed across the first resistor, if the voltmeter resistance is 143000 ohms?
Not sure where to go with this problem, Its easy to get the current for the series but im not sure how the voltmeter resistance factors in.
what part of the current is being shunted through the meter?
143/(143+654)
That is the error.
this didn't seem to work. do the voltmeter and resistor have to be treated as a parallel connection?
Given:
E = 65 Volts.
R1 = 65,400 Ohms.
R2 = 37,700Ohms.
Rm = 143,000 Ohms.
Voltmeter disconnected:
I = E/(R1+R2) = 65/(65,400+37,700) = 0.00063A.
V1 = I*R1 = 0.00063 * 65,400 = 41.23 Volts.
Voltmeter connected:
1/R = 1/R1 + 1/Rm = 1/65,400+1/143,000,
R = 44,883.3 Ohms.
I = E/(R+R2) = 65/(44,883.3+37,700) = 0.000787A.
V1 = I*R = 0.000787 * 44,883.3 = 35.33 Volts.
%Error = ((41.23-35.33)/41.23) * 100% = 14.3.
To solve this problem, we need to use Ohm's Law and the concept of voltage dividers.
Let's start by finding the total resistance (Rt) of the circuit. Since the two resistors are connected in series, we can simply add their values:
Rt = 65400 + 37700 = 103100 ohms
Next, we can use Ohm's Law (V = I * R) to find the current (I) in the circuit. The voltage (V) is given as 65 V, and the total resistance (Rt) is 103100 ohms:
I = V / Rt = 65 V / 103100 ohms = 0.00063 A
Now, let's consider the voltmeter resistance (Rv) and how it affects the voltage measurement. When the voltmeter is connected in parallel to one of the resistors, it creates a voltage divider circuit. The voltage across the resistor (Vr) can be calculated using the formula:
Vr = V * (Rv / (Rv + R))
Where V is the battery voltage, Rv is the voltmeter resistance, and R is the resistor being measured. In this case, we want to find the voltage across the first resistor (V1) given its resistance (R1 = 65400 ohms).
V1 = V * (Rv / (Rv + R1))
Substituting the given values, we get:
V1 = 65 V * (143000 ohms / (143000 ohms + 65400 ohms)) ≈ 35.11 V
Now, let's calculate the theoretical voltage (Vth) across the first resistor using the known value of the current (I) and resistance (R1):
Vth = I * R1 = 0.00063 A * 65400 ohms ≈ 40.982 V
Now, we can calculate the percentage error in the voltmeter reading by comparing the measured voltage (V1) to the theoretical voltage (Vth):
% Error = |(V1 - Vth) / Vth| * 100
% Error = |(35.11 V - 40.982 V) / 40.982 V| * 100 ≈ 14.34%
Therefore, the percent error in the voltmeter reading when placed across the first resistor is approximately 14.34%.