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Physics E&M
Posted 4 months ago

17 Several e.m.f.s, resistors and a switch, SS, are connected as shown in the circuit diagram below.
I1,I2I_{1}, I_{2} and I3I_{3} show the current flow in the circuit. Determine the value of I2I_{2}
(a) when SS is opened
[4]
(b) when SS is closed
[3][3]
18 (a) Three identical capacitors are connected to an e.m.f., EE, as shown in the circuit diagram below.
(i) Derive an expression for the effective capacitance in terms of CC.
[3]
(ii) Determine the total energy stored in terms of CC and EE.
[2]
(b) A fully charged parallel plate capacitor of capacitance 100μF100 \mu \mathrm{F}, a variable resistor, RR, and 'a switch, SS, are connected as shown in the circuit diagram below.
(i) When SS is closed, the charge reduces to half in time of 3.0 ms . Determine the resistance of RR.
[4]
(ii) If a material of dielectric constant 4.2 is inserted in between the plates, determine the time constant.
[2]
(c) Describe the effect of dielectric on the capacitance when a capacitor is connected to a potential difference.
[4][4]
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Answer from Sia
Posted 4 months ago
Sure, let's break down the questions one by one.

 Question 1:
Determine the value of I2I_{2}:
(a) when SS is opened
(b) when SS is closed

# Answer:
For part (a):
D


For part (b):
D


# Explanation:
Key Concept
Kirchhoff's Voltage Law (KVL)
Explanation
When the switch SS is open, the circuit is divided into separate loops, and we can apply KVL to each loop to find the current I2I_2. When the switch SS is closed, the circuit forms a single loop, and we again use KVL to find the new value of I2I_2.


 Question 2:
(a) Three identical capacitors are connected to an e.m.f., EE:
(i) Derive an expression for the effective capacitance in terms of CC.
(ii) Determine the total energy stored in terms of CC and EE.

# Answer:
For part (i):
D


For part (ii):
D


# Explanation:
Key Concept
Capacitors in Parallel
Explanation
When capacitors are connected in parallel, the total capacitance CtotalC_{total} is the sum of the individual capacitances: Ctotal=C+C+C=3CC_{total} = C + C + C = 3C. The energy stored in a capacitor is given by U=12CV2U = \frac{1}{2} C V^2. For three capacitors in parallel, the total energy stored is Utotal=12(3C)E2U_{total} = \frac{1}{2} (3C) E^2.


 Question 3:
(b) A fully charged parallel plate capacitor of capacitance 100μF100 \mu \mathrm{F}, a variable resistor, RR, and a switch, SS:
(i) When SS is closed, the charge reduces to half in time of 3.0 ms. Determine the resistance of RR.
(ii) If a material of dielectric constant 4.2 is inserted in between the plates, determine the time constant.

# Answer:
For part (i):
D


For part (ii):
D


# Explanation:
Key Concept
RC Time Constant
Explanation
The time constant τ\tau for an RC circuit is given by τ=RC\tau = RC. When the charge reduces to half, the time t=τln(2)t = \tau \ln(2). Given t=3.0t = 3.0 ms, we can solve for RR. For part (ii), inserting a dielectric increases the capacitance by a factor of the dielectric constant, so the new time constant is τ=R(4.2100μF)\tau' = R \cdot (4.2 \cdot 100 \mu \mathrm{F}).


 Question 4:
(c) Describe the effect of dielectric on the capacitance when a capacitor is connected to a potential difference.

# Answer:
D


# Explanation:
Key Concept
Dielectric Effect
Explanation
A dielectric material increases the capacitance of a capacitor by reducing the electric field within the capacitor for a given charge on the plates. This is quantified by the dielectric constant kk, where the new capacitance C=kCC' = kC.

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