Kirchoff’s Second Law

Questions :

1.

From that picture,

Known :E1 = 10 V dan r1 = 0 ΩE2 = 12 V dan r2 = 0 ΩR1 = 0,3 ΩR2 = 1,5 ΩR3 = 0,5 Ω

Determine the I1, I2, and I3 !

a. 1,48 A, 4,89A, 4,63 A

b. 1,48 A, 4,89 A, 6,73 A

c. 1,58 A, 4,89 A, 6,73 A

d. 1,58 A, 4,79 A, 6,83 A

e. 1,58 A, 4,79 A, 6,83 A

2. How much is VAB?

a. 3 V

b. 4 V

c. 5 V

d. 6 V

e. 7V

3. Determine I1, I2, and I3 !

a. 4,53 A, 0,88 A, -3,65 A

b. 4,53 A, 0,88 A, 3,65 A

c. 4,63 A, 0,98 A, -3,65 A

d. 4,63 A, 0,98 A, 3,65 A

e. 4,73 A, 0,98 A, -3,65 A

4. Determine I1, I2, I3 !

If :E1 = 8 VE2 = 18 VR1 = 4 ΩR2 = 2 ΩR3 = 6 Ωa.

5. Based on Kirchoff’s Second Law, determine the electric current !

6. Determine the electric current in 2 Ω resistance (AB) !

Answers :

1. I3 = I1 + I2

First Loop :∑E + ∑IR = 0- E1 + I1R1 + I3R2 = 0- 10 + 0,3 I1 + 1,5 I3 = 0 0,3 I1 + 1,5 I3 = 10

0,3 I1 + 1,5 (I1 + I2) = 100,3 I1 + 1,5 I1 + 1,5 I2 = 101,8 I1 + 1,5 I2 = 10 (x2)3,6 I1 + 3 I2 = 20 … (1)

Second Loop :∑E + ∑IR = 0- E2 + I2R3 + I3R2 = 0- 12 + 0,5 I2 + 1,5 I3 = 0 0,5 I2 + 1,5 I3 = 120,5 I2 + 1,5 (I1 + I2) = 120,5 I2 + 1,5 I1 + 1,5I2 = 121,5 I1 + 2 I2 = 12 (x1,5)2,25 I1 + 3 I2 = 18

3,6 I1 + 3 I2 = 202,25 I1 + 3 I2 = 18----------------------- -1,35 I1 = 2I1 = 1,48 A

3 I2 = 20 – 3,6 I1

3 I2 = 20 – 3,6 (1,48)3 I2 = 20 – 5,3283 I2 = 14, 67 AI2 = 4,89 A

I3 = I1 + I2

I3 = 1,48 + 4,89I3 = 6,37 A

2. VAB – 10 + 2 – (-4)= 0 VAB – 4 = 0

VAB = 4V

3. I2 = I1 + I3

First Loop : ∑E + ∑IR = 0-E1 + I1R1 - I2R2 = 0-20 + 5 I1 - 3 I2 = 05 I1 - 3 I2 = 205 I1 – 3 (I1 + I3) = 205 I1 – 3 I1 – 3 I3 = 202 I1 – 3 I3 = 20 (x3)6 I1 – 9 I3 = 60 … (1)

Second Loop : ∑E + ∑IR = 0E2 + I2R2 + I3R3 = 012 + 3 I2 + 4 I3 = 03 I2 + 4 I3 = -123 (I1 + I3) + 4 I3 = -123 I1 + 3 I3 + 4 I3 = -123 I1 + 7 I3 = -12 (x2)6 I1 + 14 I3 = -24

6 I1 + 14 I3 = -246 I1 – 9 I3 = 60----------------------- -23 I3 = -84I3 = - 3,65 A

6 I1 – 9 I3 = 606 I1 = 60 + 9 I3

6 I1 = 60 + 9 (- 3,65)6 I1 = 60 – 32,856 I1 = 27,15I1 = 4,53 A

I2 = I1 + I3

I2 = 4,53 -3,65I2 = 0,88 A

4. I3 = I1 + I2

First Loop :∑E + ∑IR = 0-E1 + I1R1 + I3R3 = 0-8 + 4 I1 + 6 I3 = 04 I1 + 6 I3 = 84 I1 + 6 (I1 + I2) = 84 I1 + 6 I1 + 6 I2 = 810 I1 + 6 I2 = 8 (x3)30 I1 + 18 I2 = 24 … (1)

Second Loop : ∑E + ∑IR = 0 -E2 + I2R2 + I3R3 = 0-18 + 2 I2 + 6 I3 = 02 I2 + 6 I3 = 182 I2 + 6 (I1 + I2) = 18

2 I2 + 6 I1 + 6 I2 = 186 I1 + 8 I2 = 18 (x5)30 I1 + 40 I2 = 90 … (2)

30 I1 + 40 I2 = 9030 I1 + 18 I2 = 24----------------------- -22 I2 = 66I2 = 3 A

6 I1 + 8 I2 = 186 I1 = 18 - 8 I2 6 I1 = 18 – 8 (3)6 I1 = -6I1 = -1 A

I3 = I1 + I2

I3 = -1 + 3I3 = 2 A

5. ∑E + ∑IR = 0E3 – E2 + E1 – I R1 – I R2 = 014 – 20 + 24 – 3I – 6I = 0-9I = -18I = 2 A

6. Electric current AB = II = I1 + I2

First Loop (ACDB) :∑E + ∑IR = 0-10 + I1 (7 +1) + I (2+0) = 0-10 + 8 I1 + 2 I = 08 I1 + 2 I = 108 I1 + 2 (I1 + I2) = 108 I1 + 2 I1 + 2 I2 = 1010 I1 + 2 I2 = 10 … (1)

Second Loop (AEFB) :∑E + ∑IR = 0-10 + I2 (7+1) + I (2+0) = 0-10 + 8 I2 + 2 I =08 I2 + 2 I = 108 I2 + 2 (I1 + I2) = 108 I2 + 2 I1 + 2 I2 = 10 10 I2 + 2 I1 = 10 (x5)

10 I1 + 50 I2 = 50 … (2)

10 I1 + 50 I2 = 5010 I1 + 2 I2 = 10----------------------- -48 I2 = 40I2 = 0,83 A

10 I1 + 2 I2 =1010 I1 = 10 – 2 I2

10 I1 = 10 – 2 (0,83)10 I1 = 8,34I1 = 0,83 A

I = I1 + I2

I = 0,83 + 0,83I = 1,66 A

Electric Energy

Questions :

1. An electric kettle 400 W/220 V is used to heat 1 kg water which c=4200J/kg˚C in 20˚C. How is the water’s temperature after being heated for 2 minutes?

2. A heater 400 W/220 V is heating 1 kg water which c=4200 J/kg˚C in 20˚C, how long it will take to boil the water?

3. An electric stove 110 V with resistance 20 Ω has been used to heat 1 kg water which c=4200 J/kg˚C in 20˚C for 5 minutes. If the stove use 110 V, how is the final temperature of the water?

4. A lamp 40 W/110V has been turning on for 10 minutes. How much is electric curent and electric energy is needed?

5. A 20 Ω resistance with the electric current 3 A. In a half minute, how much energy which is lost from the resistance?

Answers :

1. P = 400 WV = 220 V

t = 2 minutes = 120 secondm = 1 kgc = 4200J/kg˚Cto = 20˚C

W = QP . t = m . c . ∆t∆t = P. t / m . c∆t = 400 . 120 / 1 . 4200∆t = 11,43˚C

∆t = tf – to

tf = ∆t + to

tf = 11,43˚C + 20˚Ctf = 31,43˚C

2. P = 400 WV = 220 Vm = 1 kgc = 4200J/kg˚Cto = 20˚Ctf = 100˚C (boiling point)

W = QP . t = m . c . ∆tt = m . c . ∆t / Pt = 1. 4200 . (100 - 20) / 400t = 840 second = 14 minutes

3. V = 110 VR = 20 Ωm = 1 kgto = 20˚Ct = 5 minutes = 300 second

W = QV2 . t / R = m . c . ∆t∆t = V2 . t / R . m . c∆t = 1102 . 300 / 20 . 1 . 4200∆t = 43,21˚C

∆t = tf - to

tf = ∆t + to

tf = 43,21 + 20tf = 63,21˚C

4. P = 40 WV = 110 Vt = 10 minutes = 600 second

W = P . tW = 40 . 600W = 24.000 J

W = V . I . tI = W / V . tI = 24.000 / 110 . 600I = 0,36 A

5. R = 20 ΩI = 30 At = 0,5 minutes = 30 second

W = I2 . R . tW = 302 . 20 . 30W = 5.400 J