To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
\(R=\frac { V^{ 2 } }{ P } =\frac { { (Ratedvoltage) }^{ 2 } }{ Ratedpower } \)
For 20W - 220V bulb,
\({ R }_{ 1 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =2420\Omega \)
For 100W - 220V bulb,
\({ R }_{ 2 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =484\Omega \)
Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, \(I=\frac { V }{ { R }_{ tot } } \)
R_tot (R₁ + R₂)
R_tot = (484 + 2420) \(\Omega\) = 2904 \(\Omega\)
\(I=\frac { 440V }{ 2904\Omega } \approx 0.151A\)
The voltage drop across the 20W bulb is
\(V_1=IR_1=\frac { 440V }{ 2904 }\times2420 \approx 366.6V\)
The voltage drop across the 100W bulb is
\({ V }_{ 2 }=I{ R }_{ 2 }=\frac { 440 }{ 2904 } 484\approx 73.3A\)
The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

Equivalent emf of the combination
ξ_eq = nξ = 4 x 9 = 36 V
ii) Equivalent internal resistance r_eq = nr = 4 x 0.1 = 0.4 Ω
iii) Total current \(I=\frac { n\xi }{ R+nr } \)
\(=\frac { 4\times 9 }{ 10+(4\times 0.1) } \)
\(=\frac { 4\times 9 }{ 10+0.4 } =\frac { 36 }{ 10.4 } \)
I = 3.46 A
iv) Potential difference across external resistance V = IR = 3.46 x 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.
v) Potential difference across each cell \(\frac { V }{ n } =\frac { 34.6 }{ 4 } =8.65V\)

We can denote the current that flows from 9V battery as I₁ and it splits up nto I₂ and (I₁ – I₂) at the junction E according Kirchoff’s current rule (KCR).
Now consider the loop EFCBE and apply KVR, we get
1I₂ + 3I₁ + 2I₁ = 9
5I₁ + I₂ = 9          (1)
Applying KVR to the loop EADFE, we get
3 (I₁ – I₂ ) – 1I₂ = 6
3I₁ – 4I₂ = 6           (2)
Solving equation (1) and (2), we get
I₁ = 1.83 A and I₂ = -0.13 A
It implies that the current in the 1 ohm resistor flows from F to E.

\(\frac { P }{ Q } =\frac { R }{ S } \), when the network is balanced
\(\frac { 500 }{ 800 } =\frac { x+400 }{ 1000 } \)
\(x+400=\frac { 5 }{ 8 } \times 1000\)
x + 400 = 625 
x = 625 – 400
x = 225 Ω

Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A - 0.4A + 0.6A - 0.5A + 0.7A - I = 0
0.7 A - I = 0
1.5A - 0.9A – I = 0
0.6A - I = 0
I = 0.6 A