For the circuit shown in the figure, what is the current through resistor R_2? 0
ID: 1408335 • Letter: F
Question
For the circuit shown in the figure, what is the current through resistor R_2? 0.371 A 0.071 A 0.039A 0.029 A In the circuit shown in the figure, four identical resistors labeled A to D are connected to a battery as shown. S_1 and S_2 are switches. Which of the following actions would result in the LEAST amount of current through resistor A? closing both switches closing S_1 only closing S_2 only leaving both switches open as shown Two resistors R_1 = 300 Ohms and R_2 = 200 Ohms are in parallel connected to a 12 volt battery. Which resistor is using less power? R_1 R_2 They use the same power since they have the same current flowing through them Can't tell from this infoExplanation / Answer
27 ans
We applying Kirchhoff's first law in this circuit
i1+i2=i3
Now we apply Kirchhoff's second law at loop ABEFA
We get=> -7-70i1+2=0
=>70i1=5
Therefore the current i1=0.07143A
Now we apply Kirchhoff's second law at loop BCDEB
-105(i1+i2)-140(i1+i2)+7=0
245i1+245i2=7
245(0.07143)+246i2=7
17.50035+245i2=7
245i2=-10.50035
Therefore the current i2=-0.04286A
Now we substitute current i1&i2 we get i3
i3=0.07143-0.04286=0.029A
Therefore the current pass through resistance R2 is 0.029A
(28) ans
(1) in this condition the switch's is closed
The resistances B,C,D are in parallel combination
So efficitive resistance Rp=R^3/2R^2=R/3
The Resistance Rp and A are series condition
The effective resistance Rs=R/3+R=4/3R=1.33 R
The current passing through resistance A is
=V/1.33R=0.752i.........(1)
Option B&c:
In this condition the current passes two resistances only
So the effective resistance b/w B&c or C&B we get
The effective resistance Rp=R^2/2R=R/2
The effective resistance Rp and resistance A are series condition
The effective resistance Rs=R+R/2=1.5R
The current passing through resistance A is =V/1.5R=0.67i
Option D:
In this condition the current passing through resistance C only
So the resistance C &A are in series condition
The effective resistance Rs=R+R=2R
The current passing through resistance A is =V/2R=0.5i
Finially we proved least current passing through resistance A in option D
(29) ans
In this circuit the resistance are parallel combination so the potential is same for both resistances
Therefore the power supplied in resistance R1
Power P1=V^2/R1=(12)^2/300=0.48W
The power supplied in resistance R2
P2=(12)^2/200=0.72W
The resistance R1 is using less power