Question

The ammeter shown in Figure P18.16 reads 2.90 A. Find I1, I2, and ε.
I1 = A
I2 = A
ε = V

Answer 1

For the top loop
starting from the battery
15-7i1-5(i1+i2)=0
15-7A-10A=0
A=0.882 A= i1=i2

for the second loop
starting from e
V-2*i2-5*(i2+i1)=0
v=12*0.882=10.588 V

Answer 2

AMMETER READING = 2.9 A
APPLY KVL IN THE UPPER LOOP
15 -7I₁ -5*2.9=0

OR I₁= 0.071

I₁+I_{2 =} 2.9 A

HENCE , I₂ =2.9 -0.07 = 2.83 A

APPLY KVL IN THE LOWER LOOP

ε - 2I₂ -5*2.9 =0

ε = 20.16 V

Answer 3

For the top loop:

15 - 7*i1 - 5*2.9 = 0 >>>> i1 = 0.071 A

-------------------------------------------------------------------------

i1 + i2 = 2.9 >>>> i2 = 2.9 - 0.071 >>>>> i2 =2.829 A

--------------------------------------------------------------------------

For the second loop:

ε - 2*i2 - 5*2.9 = 0 >>>> ε = 5*2.9 + 2*2.829 >>>> ε = 20.158 Volt

Answer 4

For the top loop
starting from the battery
15-7i1-5(i1+i2)=0
15-7A-5A=0
A=1.25 A= i1=i2

for the second loop
starting from e
V-2*i2-5*i2=0
v=7*1.25=8.75 V

Answer 5

Using Kirchhoff’s Rules.
At Junction A: A 00.2
21
III ==+
Loop 1 (starting from B):
0 ) A)(5.00 00.2() 00.7(V 0.15 I
1
-O- O =
A 714.0
V 00.5V 0.10V 15.0) 00.7(
1
1
=?
=-=O?
I
I
Thus:
A 29.1A 286.1A 714.0A 00.2A 00.2 I
2
I
1
-=-= = =
I
A B
Loop 1
Loop 2
Loop 2 (starting from B): e - O - O = 0) A)(5.00 00.2() A)(2.00 286.1(
e =? V 6.12
Alternative Method:
Since we know I the potential between A and B is,
IV =O=? O = V 10.0) A)(5.00 (2.00) 00.5(
Therefore the potential drop over the 7.00 O resistor is ?V7 = 15.0 V - 10.0 V = 5.00 V.
Therefore the current I1 is A 714.0
7.00
V 00.5
1
=
O
I =
Therefore since at A: I = I1 + I2,
A 29.1A 714.0A 00.2
2
III
1
-=-=? =
Therefore the potential drop across the 2.00 O resistor is
?V2 = I2(2.00 O) = (1.29 A)(2.00 O) = 2.58 V.
So the potential e = ?V2 + ?V = 2.58 V + 10.0 V = 12.6 V