Current through bulb 4 is maximum

Step 1: Resistance of the bulbs

$$R_1 = \frac{110^2}{60} \approx 202 \, \Omega, \quad R_2 = \frac{110^2}{100} = 121 \, \Omega.$$

Step 2: Total current in the circuit

In series, the same current flows through both bulbs:

$$I = \frac{V_{\text{total}}}{R_1 + R_2} = \frac{220}{202 + 121} \approx 0.682 \, \text{A}.$$

Step 3: Voltage across each bulb

• Voltage across the 60 W bulb:

$$V_1 = I \cdot R_1 = 0.682 \cdot 202 \approx 137.6 \, \text{V}.$$

• Voltage across the 100 W bulb:

$$V_2 = I \cdot R_2 = 0.682 \cdot 121 \approx 82.5 \, \text{V}.$$

Step 4: Compare with rated voltage

• The 60 W bulb experiences
  $137.6 \, \text{V}$ 

  , exceeding its $110 \, \text{V}$ 

  rating, so it fuses.

• The 100 W bulb experiences
  $82.5 \, \text{V}$ 

  , within its $110 \, \text{V}$ 

  rating.

Conclusion

The 60 W bulb will fuse.

$$\boxed{\text{Answer: 60 W bulb.}}$$

Let's go through a more detailed, step-by-step approach to solving the problem correctly, and we’ll arrive at the answer

$m = 3$

and

$n = 15$

.

Given:

• 45 cells, each with an internal resistance of 0.5Ω.
• External resistance
  $R_{\text{ext}} = 2.5 \, \Omega$ 

  .

• We want to determine the configuration that maximizes the current.

We are arranging the cells in series and parallel, so:

• 
  $m$ 

  = number of rows (parallel branches of cells)

• 
  $n$ 

  = number of cells in each row (connected in series)

Step 1: Internal resistance of one row

When

$n$

cells are connected in series, the internal resistance for each row (denoted as

$R_{\text{internal, row}}$

) is the sum of the internal resistances of each cell:

$$R_{\text{internal, row}} = n \times r_{\text{cell}} = n \times 0.5 \, \Omega$$

Step 2: Internal resistance of the entire arrangement

Since there are

$m$

rows connected in parallel, the total internal resistance

$R_{\text{internal, total}}$

of the entire setup is:

$$R_{\text{internal, total}} = \frac{R_{\text{internal, row}}}{m} = \frac{n \times 0.5}{m}$$

Step 3: Total resistance in the circuit

The total resistance in the circuit is the sum of the external resistance

$R_{\text{ext}}$

and the total internal resistance of the cells

$R_{\text{internal, total}}$

:

$$R_{\text{total}} = R_{\text{ext}} + R_{\text{internal, total}} = 2.5 + \frac{n \times 0.5}{m}$$

Step 4: Total current using Ohm's Law

The current through the circuit can be calculated using Ohm's Law,

$I = \frac{V}{R_{\text{total}}}$

, where

$V$

is the total voltage supplied by the cells.

For maximum current, we want to minimize

$R_{\text{total}}$

, which means minimizing

$R_{\text{internal, total}}$

.

Step 5: Constraint on $m$

and $n$

We are given that there are 45 cells in total, so:

$$m \times n = 45$$

Thus,

$n = \frac{45}{m}$

.

Step 6: Substituting $n = \frac{45}{m}$

into the total resistance formula

Substitute

$n = \frac{45}{m}$

into the formula for

$R_{\text{total}}$

:

$$R_{\text{total}} = 2.5 + \frac{\left(\frac{45}{m}\right) \times 0.5}{m}$$

Simplifying this:

$$R_{\text{total}} = 2.5 + \frac{22.5}{m^2}$$

Step 7: Minimizing $R_{\text{total}}$

Now, to minimize the total resistance, we need to minimize

$R_{\text{total}} = 2.5 + \frac{22.5}{m^2}$

.

Since

$\frac{22.5}{m^2}$

decreases as

$m$

increases, we need to check the values of

$m$

that are divisors of 45.

Step 8: Trying possible values of $m$

Let’s try a few possible values for

$m$

:

1. For
   $m = 3$ 

   :

$$n = \frac{45}{3} = 15$$

$$R_{\text{total}} = 2.5 + \frac{22.5}{3^2} = 2.5 + \frac{22.5}{9} = 2.5 + 2.5 = 5 \, \Omega$$

2. For
   $m = 5$ 

   :

$$n = \frac{45}{5} = 9$$

$$R_{\text{total}} = 2.5 + \frac{22.5}{5^2} = 2.5 + \frac{22.5}{25} = 2.5 + 0.9 = 3.4 \, \Omega$$

3. For
   $m = 9$ 

   :

$$n = \frac{45}{9} = 5$$

$$R_{\text{total}} = 2.5 + \frac{22.5}{9^2} = 2.5 + \frac{22.5}{81} = 2.5 + 0.277 \approx 2.78 \, \Omega$$

4. For
   $m = 15$ 

   :

$$n = \frac{45}{15} = 3$$

$$R_{\text{total}} = 2.5 + \frac{22.5}{15^2} = 2.5 + \frac{22.5}{225} = 2.5 + 0.1 = 2.6 \, \Omega$$

Step 9: Conclusion

The configuration that minimizes the total resistance and maximizes the current is when

$m = 3$

and

$n = 15$

, which results in a total resistance of 5Ω. Thus, the answer is:

$$\boxed{m = 3, \, n = 15}$$

To find the resistance

$R$

that must be placed in series with the bulb, let's analyze the problem step by step.

Given:

1. Power of the bulb (
   $P$ 

   ) = 500 W,

2. Voltage rating of the bulb (
   $V_b$ 

   ) = 100 V,

3. Supply voltage (
   $V_s$ 

   ) = 200 V.

Step 1: Resistance of the bulb

The resistance of the bulb (

$R_b$

) can be calculated using the formula:

$$R_b = \frac{V_b^2}{P}.$$

Substitute the values:

$$R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega.$$

Step 2: Total current through the circuit

The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:

$$I = \frac{P}{V_b}.$$

Substitute the values:

$$I = \frac{500}{100} = 5 \, \text{A}.$$

Step 3: Voltage drop across the series resistor

The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor

$R$

is:

$$V_R = V_s - V_b.$$

Substitute the values:

$$V_R = 200 - 100 = 100 \, \text{V}.$$

Step 4: Resistance of the series resistor

Using Ohm's law, the resistance of the series resistor is:

$$R = \frac{V_R}{I}.$$

Substitute the values:

$$R = \frac{100}{5} = 20 \, \Omega.$$

Final Answer:

The resistance that must be placed in series with the bulb is:

$$\boxed{20 \, \Omega}.$$