Two resistors of resistances `R_1 = 100 pm 3` ohm and `R_2 = 200 pm 4 ` ohm are connected (a) in series , (b) in parrallel. Find the equivalent resistance of the (a) series combination , (b) parallel combination . Use for (a) the relation `R = R_1 + R_2 and ` for (b) `1/R = 1/(R_1) + 1/(R_2) and (DeltaR')/(R^(,2)) = (DeltaR_1)/(R_1^2) + (DeltaR_2)/(R_2^2)`

Options

A. `300 Omega`
B. `(300 pm 7)Omega`
C. `(300 pm 3)Omega`
D. `(300 pm 4)Omega`

Answer

To solve the problem, we will calculate the equivalent resistance of two resistors connected in series and parallel, taking into account their uncertainties.

### Given:
- Resistor 1: \( R_1 = 100 \pm 3 \, \Omega \)
- Resistor 2: \( R_2 = 200 \pm 4 \, \Omega \)

### (a) Series Combination
The formula for equivalent resistance in series is:
\[ R = R_1 + R_2 \]

**Step 1: Calculate the equivalent resistance**
\[ R = 100 + 200 = 300 \, \Omega \]

**Step 2: Calculate the uncertainty in the equivalent resistance**
The uncertainties in series add up:
\[ \Delta R = \Delta R_1 + \Delta R_2 = 3 + 4 = 7 \, \Omega \]

**Final Result for Series Combination:**
\[ R_{\text{series}} = 300 \pm 7 \, \Omega \]

### (b) Parallel Combination
The formula for equivalent resistance in parallel is:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]

**Step 1: Calculate the equivalent resistance**
\[ \frac{1}{R_p} = \frac{1}{100} + \frac{1}{200} \]
Calculating this gives:
\[ \frac{1}{R_p} = \frac{2}{200} + \frac{1}{200} = \frac{3}{200} \]
Thus,
\[ R_p = \frac{200}{3} \approx 66.67 \, \Omega \]

**Step 2: Calculate the uncertainty in the equivalent resistance**
Using the formula for the uncertainty in parallel resistors:
\[ \frac{\Delta R'}{R'^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \]
We can rearrange this to find \(\Delta R'\):
\[ \Delta R' = R'^2 \left( \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \right) \]
Substituting the values:
- \( R' = \frac{200}{3} \)
- \( \Delta R_1 = 3 \)
- \( R_1 = 100 \)
- \( \Delta R_2 = 4 \)
- \( R_2 = 200 \)

Calculating:
\[ \Delta R' = \left( \frac{200}{3} \right)^2 \left( \frac{3}{100^2} + \frac{4}{200^2} \right) \]
Calculating each part:
\[ \left( \frac{200}{3} \right)^2 = \frac{40000}{9} \]
\[ \frac{3}{10000} + \frac{4}{40000} = \frac{3 \times 4 + 4}{40000} = \frac{12 + 4}{40000} = \frac{16}{40000} = \frac{1}{2500} \]
Thus,
\[ \Delta R' = \frac{40000}{9} \times \frac{1}{2500} = \frac{16}{9} \approx 1.78 \, \Omega \]

**Final Result for Parallel Combination:**
\[ R_{\text{parallel}} = 66.67 \pm 1.78 \, \Omega \]

### Summary of Results:
- Equivalent resistance in series: \( R_{\text{series}} = 300 \pm 7 \, \Omega \)
- Equivalent resistance in parallel: \( R_{\text{parallel}} = 66.67 \pm 1.78 \, \Omega \)