The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.
The paths AA', AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further,  at the corners AA', B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC'EA, and apply Kirchhoff’s second rule:
–IR – (1/2)IR – IR + \(\epsilon\) = 0
where R is the resistance of each edge and e the emf of battery. Thus,
\(\epsilon = \frac{5}{2}IR\)
The equivalent resistance R_eq of the network is
\(R_{eq}= \frac{\epsilon}{3I}=\frac{5}{6}R\)
For R = 1 Ω, R_eq = (5/6) Ω and for e = 10 V, the total current (= 3I ) in
the network is
3I = 10 V/(5/6) Ω = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the figure.

Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
\(B=\frac{\mu_{0} N I}{l}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}\)
\(=8 \pi \times 10^{-3}=2.512 \times 10^{-2} T\)
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10^–2 T.

Current flowing through various branches of the circuit is represented in the given figure.

I₁ = Current flowing through the outer circuit
I₂ = Current flowing through branch AB
I₃ = Current flowing through branch AD
I₂ - I₄ = Current flowing through branch BC
I₃ + I₄ = Current flowing through branch CD
I₄ = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I₂ + 5I₄ - 5I₃ = 0
2I₂ + I₄ - I₃ = 0
I₃ = 2I₂ + I₄ … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I₂ - I₄) - 10(I₃ + I₄) - 5I₄ = 0
5I₂ + 5I_{4 -} 10I₃ - 10I₄ - 5I₄ = 0
5I₂ - 10I₃ - 20I₄ = 0
I₂ = 2I₃ + 4I₄ … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
-10 + 10 (I₁) + 10(I₂) + 5(I₂ - I₄) = 0
10 = 15I₂ + 10I₁ - 5I₄
3I₂ + 2I₁ - I₄ = 2 … (3)
From equations (1) and (2), we obtain
I₃ = 2(2I₃ + 4I₄) + I₄
I₃ = 4I₃ + 8I₄ + I4
- 3I₃ = 9I₄
- 3I₄ = + I₃ … (4)
Putting equation (4) in equation (1), we obtain
I₃ = 2I₂ + I₄
- 4I₄ = 2I₂
I₂ = - 2I₄ … (5)
It is evident from the given figure that,
I₁ = I₃ + I₂ … (6)
Putting equation (6) in equation (1), we obtain
3I₂ +2(I₃ + I₂) - I₄ = 2
5I₂ + 2I₃ - I₄ = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(- 2 I₄) + 2(- 3 I₄) - I₄ = 2
- 10I₄ - 6I₄ - I₄ = 2
17I₄ = - 2
\(I_{4}=-\frac{2}{17} A\)
Equation (4) reduces to
I₃ = - 3(I₄)
\(=-3\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{2}=-2\left(I_{4}\right)\)
\(=-2\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{2}-I_{4}=\frac{4}{17}-\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{3}+I_{4}=\frac{6}{17}+\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{1}=I_{3}+I_{2}\)
\(=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} A\)
Therefore, current in branch \(A B=\frac{4}{17} A\)
In branch BC \(=\frac{6}{17} A\)
In branch CD = \(-\frac{4}{17} A\)
In branch AD = \(\frac{6}{17} A\)
In branch BD = \(\left(-\frac{2}{17}\right) A\)
Total current = \(\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} A\)

Considering the mesh BADB, we have
100I₁ + 15I_g - 60I₂ = 0
or 20I₁ + 3I_g - 12I₂ = 0
Considering the mesh BCDB, we have
10 (I₁ - I_g) - 15I_g - 5 (I₂ + I_g) = 0
10I₁ - 30I_g - 5I₂ = 0
2I₁ - 6I_g - I₂ = 0
Considering the mesh ADCEA,
60I₂ + 5 (I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2
Multiplying by 10
20I₁ - 60I_g - 10I₂ = 0
From we have
63I_g - 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into we get
13 (31.5I_g ) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA

Given that the Inductance of the inductor in the circuit is, L = 5.0 H
Given that the Capacitance of the capacitor in the circuit is , C = 80 μH = 80 x 10 ^- 6 F
Given that Resistance of the resistor in the circuit, R = 40 Ω
Value of Potential of the variable voltage supply, V = 230 V
(a) We know that the Resonance angular frequency can be obtained by the following relation :
\(\omega_{r}=\frac{1}{\sqrt{L C}} \omega_{r}=\frac{1}{\sqrt{5 x 80 x 10-6}} \omega_{r}=\frac{10^{3}}{20}=50 \mathrm{rad} / \mathrm{sec}\)
Thus, the circuit encounters resonance at a frequency of 50 rad/s.
(b) We know that the Impedance of the circuit can be calculated by the following relation :
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
At resonant condition,
X _L = X _C
Z = R = 40 Ω
At resonating frequency amplitude of the current can be given by the following relation :
\(I_{0}=\frac{V_{0}}{Z}\)
where,
V ₀ = peak voltage = \(\sqrt{2} V\)
Therefore,
\(I_{0}=\frac{\sqrt{2 V}}{Z}=\frac{\sqrt{2} \times 230}{40}=8.13 \mathrm{~A}\)
Thus, at resonant condition, the impedance of the circuit is calculated to be 40 Ω and the amplitude of the current is found to be  8.13 A
c) rms potential drop across the inductor in the circuit,
( V _L ) _rms = I x ω _r L
Where,
\(I_{r m s}=\frac{I_{0}}{\sqrt{2}}=\frac{\sqrt{2} V}{\sqrt{2} Z}=\frac{230}{40}=\frac{23}{4} A\)
Therefore, ( V _L ) _rms
\(\frac{23}{4} \times 50 \times 5=1437.5 \mathrm{~V}\)
We know that the Potential drop across the capacitor can be calculated with the following relation :
\(\left(V_{c}\right)_{r m s}=I \times \frac{1}{\omega_{r} C}=\frac{23}{4} \times \frac{1}{50 \times 80 \times 10^{-6}}=1437.5 V\)
We know that the Potential drop across the resistor can be calculated with the following relation :
\(\left(V_{R}\right)_{r m s}=I R=\frac{23}{4} \times 40=230 \mathrm{~V}\)
Now, Potential drop across the LC connection can be obtained by the following relation :
V _{L C}  = I ( X _L − X _C )
At resonant condition,
X _L = X _C
V _{L C} = 0
Therefore, it has been proved from the above equation that the potential drop across the LC connection is equal to zero at a frequency at which resonance occurs.

The supply frequency and the natural frequency are equal at resonance condition in the circuit.
Given Resistance of the resistor, R = 20 Ω
Given Inductance of the inductor, L = 1.5 H
Given Capacitance of the capacitor , C = 35 μF = 30 x 10 ^- 6 F
An AC source with a voltage of V = 200 V is connected to the LCR circuit,
We know that the Impedance of the above combination can be calculated by the following relation,
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
At resonant condition in the circuit , X _L = X _C
Therefore , Z = R = 20 Ω
We know that Current in the network is given by the relation :
\(I=\frac{V}{Z}=\frac{200}{20}=10 A\)
Therefore, the average power that is being transferred to the circuit in one full cycle :
V I = 200 x 10 = 2000 W

a) Given: The peak voltage of supply is 100 V.
The rms voltage is give as,
v m = 2 ×V
Where, the peak value of supply voltage is v m and its rms value is V.
By substituting the given values in the above equation, we get
300= 2 ×V V= 300 2 =212.1 V
Thus, the value of rms voltage is 212.1V.
b) Given: The rms current in an ac circuit is 10 A.
The peak current in the circuit is given as,i m = 2 ×I
Where, the peak current in an ac circuit is i m and its rms value is I.
By substituting the given values in the above equation, we get
i m = 2 ×10 =14.1 A
Thus, the value of peak current in the given ac circuit is 14.1 A.

Here, R = 0.8 cm = 8 x 10^-3 m ;
I = 100 A
\((a) \ { B }_{ inside }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2Ir }{ { R }^{ 2 } } \) 
\(={ 10 }^{ -7 }\times \frac { 2\times 100\times { \left( 0.2\times { 10 }^{ -2 } \right) }^{ 2 } }{ { \left( 8\times { 10 }^{ -3 } \right) }^{ 2 } } \) 
= 6.25 x 10^-4 T
\((b) \ { B }_{ surface }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ { R } } ={ 10 }^{ -7 }\times \frac { 2\times 100 }{ 8\times { 10 }^{ -3 } } \) 
= 2.5 x 10^-5 T
(c) When point is outside the wire,
r = 0.8 + 0.4 = 1.2 cm = 1.2 x 10^-2 m.
\({ B }_{ outside }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ { r } } ={ 10 }^{ -7 }\times \frac { 2\times 100 }{ 1.2\times { 10 }^{ -2 } } \) 
= 1.67 x 10^-5 T.

Magnetic field strength, B = 6.5 G = 6.5 x 10^–4 T
Speed of the electron, v = 4.8 x 10⁶ m/s
Charge on the electron, e = 1.6 x 10^–19 C
Mass of the electron, m_e = 9.1 x 10^–31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sinθ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
\(F_{c}=\frac{m v^{2}}{r}\)
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
F_C = F
\(\frac{m v^{2}}{r}=e v B \sin \theta\)
\(r=\frac{m v}{B e \sin \theta}\)
\(=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{6.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times \sin 90^{\circ}}\)
= 4.2 x 10 ^-2 m = 4.2 cm
Hence, the radius of the circular orbit of the electron is 4.2 cm.

Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V¹
R is connected to the storage battery in series. Hence, it can be written as
V¹ = V - E
V¹ = 120 - 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
\(I=\frac{V^{1}}{R+r}\)
\(=\frac{112}{15.5+5}=\frac{112}{16}=7 A\)
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 - 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.