( a ) A voltmeter and an ammeter can be connected as shown in Fig. 26–70a to measure a resistance R . If V is the voltmeter reading, and I is the ammeter reading, the value of R will not quite be V / I (as in Ohm’s law) because some of the current actually goes through the voltmeter. Show that the actual value of R is given by 1 R = I V − 1 R V . where R V is the voltmeter resistance. Note that R ≈ V / I if R V ≫ R. ( b ) A voltmeter and an ammeter can also be connected as shown in Fig. 26–70b to measure a resistance R . Show in this case that R = V I − R A , where V and I are the voltmeter and ammeter readings and R A is the resistance of the ammeter. Note that R ≈ V / I if R A ≪ R . FIGURE 26–70 Problem 81.
( a ) A voltmeter and an ammeter can be connected as shown in Fig. 26–70a to measure a resistance R . If V is the voltmeter reading, and I is the ammeter reading, the value of R will not quite be V / I (as in Ohm’s law) because some of the current actually goes through the voltmeter. Show that the actual value of R is given by 1 R = I V − 1 R V . where R V is the voltmeter resistance. Note that R ≈ V / I if R V ≫ R. ( b ) A voltmeter and an ammeter can also be connected as shown in Fig. 26–70b to measure a resistance R . Show in this case that R = V I − R A , where V and I are the voltmeter and ammeter readings and R A is the resistance of the ammeter. Note that R ≈ V / I if R A ≪ R . FIGURE 26–70 Problem 81.
(a) A voltmeter and an ammeter can be connected as shown in Fig. 26–70a to measure a resistance R. If V is the voltmeter reading, and I is the ammeter reading, the value of R will not quite be V/I (as in Ohm’s law) because some of the current actually goes through the voltmeter. Show that the actual value of R is given by
1
R
=
I
V
−
1
R
V
.
where RV is the voltmeter resistance. Note that R ≈ V/I if RV ≫ R. (b) A voltmeter and an ammeter can also be connected as shown in Fig. 26–70b to measure a resistance R. Show in this case that
R
=
V
I
−
R
A
,
where V and I are the voltmeter and ammeter readings and RA is the resistance of the ammeter. Note that R ≈ V/I if RA ≪ R.
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During an experiment to verify Ohm's law, the voltage supplied and the current through a circuit are measured.
[Voltage is measured in Volt (V) and current in Ampere (A)].
ww
R
on
Battery
The measured value of the current is I = 3.1 ± 0.2 A and that of the voltage is V = 14 0.5 V. The resistance of
the circuit (in N) can be calculated using the formula, R = V/I,
Calculate the,
a) Resistance (in 2) =
b) Fractional uncertainty in the resistance =
c) Absolute uncertainty (in 2) in the resistance=
6. (a) Charge Q coulombs at time t seconds
is given by the differential equation
Õp
dQ Q
RE+
=0, where C is the capaci-
dt
tance in farads and R the resistance in
ohms. Solve the equation for Q given
that Q=Qo when t=0.
(b) A circuit possesses a resistance of
250 × 10³ 2 and a capacitance of
8.5 x 10–6F, and after 0.32 seconds
the charge falls to 8.0C. Determine
the initial charge and the charge after
1 second, each correct to 3 significant
figures.
-t
(b) Determine the voltage across resistor 2.5 Q , v (show full calculation)
60 Q
12 Q
+ v -
2.5 Q
80 Q
40 V
15 Q
20 2
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
How To Solve Any Resistors In Series and Parallel Combination Circuit Problems in Physics; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=eFlJy0cPbsY;License: Standard YouTube License, CC-BY