Ch27

 

𝜏=𝑅𝐶 𝑉=𝜀=𝑞0𝐶

During the charging:kirchhoff's laws:

Ch27Q.19.A total resistance of 3.00 is to be produced by connecting an unknown resistance to a 12.0 resistance. (a) What must be the value of the unknown resistance?(b) Should it be connected in series or in parallel?

Q.23.In Fig. 27-35, R1 = 100, R2 =50, and the ideal batteries have emfs ɛ1= 6.0 V, ɛ2= 5.0 V, and ɛ3= 4.0 V. Find (a) the current in resistor1(b) The current in resistor 2,(c) The potential difference between points a and b.

Ch27Q.58.In an RC series circuit, ɛ= 12.0 V, resistance R= 1.40 M Ω, and capacitance C= 1.80 µF.(a) Calculate the time constant.(b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 16.0 µC?

Q.61.A 15.0 k Ω resistor and a capacitor are connected in series, and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 5.00 V in 1.30 µs. (a) Calculate the time constant of the circuit.(b)Find the capacitance of the capacitor.

1) labeling junctions in the circuit (the junction is any point in a circuit where three or more wires meet).

2) labeling currents direction and applying Kirchhoff’s Current Law (KCL):Which states that, the total electric current entering any junction must equal the total electric current leaving that junction.(Direction of currents will be confirmed once we complete the problem).

𝑅2

𝜺𝟏

𝜺𝟐 𝜺𝟑

𝑅1

DA

𝐼 3

𝐼 1𝐼 2

𝐼 3

𝐼 1𝐼 2

𝐼 1+ 𝐼3=𝐼 2Node (D) :

𝐼 1+ 𝐼2=𝐼 3Node (A) :

𝐼 2=𝐼 1+𝐼 3Node (A) :

They should be the same

They are the same

4) Labeling closed loops in the circuit.

3) choose loops direction (either clockwise or counterclockwise)

5) Applying Kirchhoff’s Voltage Law (KVL(:Which states that the summation of all voltage drops in a closed loop must equal to zero.

Ԑ is positive, when a charge moves from the negative terminal to the positive terminal in a battery ( from low to high potential and vice versa. I is positive, when the direction of the current and loop are the same and vice versa.

𝑅2

𝜺𝟏

𝜺𝟐 𝜺𝟑

𝑅1

𝐼 2

𝐼 3

𝐼 1

DA

𝐼 3

𝐼 1𝐼 2

F

G

E

BC

EFGADE

DABCD

𝐼 1𝐼 1

𝐼 1

𝐼 3

−𝜀1− 𝐼1𝑅2+𝜀3+𝜀2=0

−𝜀2+𝐼 3𝑅1=0

Note that the negative sign of the current indicates that the assumed direction we chose is the opposite of the actual direction the current is flowing in.

𝑅2

𝜺𝟏

𝜺𝟐 𝜺𝟑

𝑅1

𝐼 2

𝐼 3

𝐼 1

DA

𝐼 3

𝐼 1𝐼 2

F

G

E

BC

𝐼 1𝐼 1

𝐼 1

𝐼 3

EFGADE

−𝜀1− 𝐼1𝑅2+𝜀3+𝜀2=0DABCD

−𝜀2+𝐼 3𝑅1=0

𝑅1=100Ω ,𝑅2=50Ω ,𝜀1=6V ,𝜀2=5V ,𝜀3=4V ,

a)

b)

c) ??¿5+4=9V OR

12 V

5 V

a)

𝑉 𝐶=𝜀 (1−𝑒−𝑡𝑅𝐶 )

⇒1−

𝑉 𝐶

𝜀=𝑒

− 𝑡𝑅𝐶

⇒𝑙𝑛(1−𝑉 𝐶

𝜀 )=− 𝑡𝑅𝐶⇒

𝑅𝐶=𝜏=−𝑡

𝑙𝑛(1−𝑉 𝐶

𝜀 )¿− 1.3×10

−6

𝑙𝑛(1− 512 )=2 .4×10− 6𝑠𝑒𝑐=3 .4 µ s

b)

𝜏=𝑅𝐶

𝜀

𝑉 𝑐

𝐶 𝑅