Q 14 please Wheatstone Bridge:(910); In the circuit shown. What is theequivalent resistance of the circuit?a) 20 b) 60c) 40d) 120Kirchhoff's LawsThe ammeter shown in the circuit reads 1.6 A(Q13): Calculate the currents 11 and 12 as (11,12) ?a) (0.3,1.3)A b) (1,0.6)Ac) (0.6,1)AGalvanometer(011); In the shown figure, a galvanometer with (full scale 5 mA, R=1 KO).The value of Rx (in k0) that makes the galvanometer reads a voltage oforder of 30 V is:460(Q14); Calculate e?a) 9.2V10 2Rsb) 8.4V c) 10VEa) 2b) 10c) 15 d) 12Rx(Q12): To convert the Galvanometer into an ammeter we must connect aa)parallel small resistor b) parallel large resistor c) series small resistor d) parallel largeresistor100d) 12VPower TransferFor the circuit shown, plot of PL, the power dissipated in the loadResistance RL, is given as a function of RL.50(Q15); The value of resistance Rs (in ) is:b)50c)80a) 30d) 40(Q16): Calculate the emf of the battery E (in Volt)?a) 8.34c) 3.09d) 2.76b) 6.0120d) (1.3,0.3)AAllt25V407.000WPL (mW)10201203.000WW2.00015.0V

Answered: Image /qna-images/answer/0fbea055-96ae-4c8c-9e92-2d46b87e4bf2.jpg

Kirchhoff's Laws The ammeter shown in the circuit reads 1.6 A (Q13): Calculate the currents 11 and 12 as (11,12) ? a) (0.3,1.3)A b) (1,0.6)A c) (0.6,1)A (Q14); Calculate ? a) 9.2V b) 8.4V c) 10V d) 12V d) (1.3,0.3)A 1.000 W 3.000 W www ILOV

Kirchhoff's Laws The ammeter shown in the circuit reads 1.6 A (Q13): Calculate the currents 11 and 12 as (11,12) ? a) (0.3,1.3)A b) (1,0.6)A c) (0.6,1)A (Q14); Calculate ? a) 9.2V b) 8.4V c) 10V d) 12V d) (1.3,0.3)A 1.000 W 3.000 W www ILOV

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter10: Direct-current Circuits

Section: Chapter Questions

Problem 69AP: Consider the circuit below. The battery has an emf of = 30.00 V and an internal resistance of r =...

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Question

Q 14 please

Wheatstone Bridge:
(910); In the circuit shown. What is the
equivalent resistance of the circuit?
a) 20 b) 60
c) 40
d) 120
Kirchhoff's Laws
The ammeter shown in the circuit reads 1.6 A
(Q13): Calculate the currents 11 and 12 as (11,12) ?
a) (0.3,1.3)A b) (1,0.6)A
c) (0.6,1)A
Galvanometer
(011); In the shown figure, a galvanometer with (full scale 5 mA, R=1 KO).
The value of Rx (in k0) that makes the galvanometer reads a voltage of
order of 30 V is:
460
(Q14); Calculate e?
a) 9.2V
10 2
Rs
b) 8.4V c) 10V
E
a) 2
b) 10
c) 15 d) 12
Rx
(Q12): To convert the Galvanometer into an ammeter we must connect a
a)parallel small resistor b) parallel large resistor c) series small resistor d) parallel large
resistor
100
d) 12V
Power Transfer
For the circuit shown, plot of PL, the power dissipated in the load
Resistance RL, is given as a function of RL.
50
(Q15); The value of resistance Rs (in ) is:
b)50
c)80
a) 30
d) 40
(Q16): Calculate the emf of the battery E (in Volt)?
a) 8.34
c) 3.09
d) 2.76
b) 6.01
20
d) (1.3,0.3)A
Allt
25V
40
7.000
W
PL (mW)
10
201
20
3.000
W
W
2.000
15.0V

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