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Step-by-step Solution
Step 1: Understand the Circuit Configuration
There are two batteries in series, each of emf 10\,\text{V} but with different internal resistances 20\,\Omega and 5\,\Omega . This series combination is connected to a parallel arrangement of two resistors 30\,\Omega and R\,\Omega .
Step 2: Write the Condition for Zero Terminal Voltage across the First Battery
We are told that the voltage difference (terminal voltage) across the battery having internal resistance 20\,\Omega is zero. For a battery with emf E and internal resistance r , its terminal potential difference is
V_\text{terminal} = E - I \, r.
If V_\text{terminal} = 0 for the first battery ( E = 10\,\text{V} , r = 20\,\Omega ), then:
0 = 10 - I \times 20.
Solving for I (total current in the circuit):
I = \frac{10}{20} = 0.5\,\text{A}.
Step 3: Determine Voltage across the Parallel Combination
Since the batteries are in series with total emf 10 + 10 = 20\,\text{V} , and we have found I = 0.5\,\text{A} . However, the first battery drops its entire emf across its own internal resistance and contributes zero net terminal voltage. Therefore, the second battery (emf 10\,\text{V} and internal resistance 5\,\Omega ) effectively drives the same current I = 0.5\,\text{A} .
Hence, the terminal voltage of the second battery is:
V_{\text{across second battery}} = 10 - I \times 5 = 10 - (0.5)\times 5 = 10 - 2.5 = 7.5\,\text{V}.
This 7.5\,\text{V} will appear across the parallel combination of 30\,\Omega and R\,\Omega .
Step 4: Relate the Voltage to the Parallel Combination
The total current through the parallel combination is 0.5\,\text{A} . The voltage across the parallel branch is 7.5\,\text{V} . Therefore, the effective resistance of the parallel combination is:
R_{\text{parallel}} = \frac{7.5\,\text{V}}{0.5\,\text{A}} = 15\,\Omega.
Step 5: Solve for R
For two resistors 30\,\Omega and R\,\Omega in parallel, the effective resistance is:
R_{\text{parallel}} = \frac{30 \times R}{30 + R}.
We know R_{\text{parallel}} = 15\,\Omega , so:
\frac{30 \times R}{30 + R} = 15.
Cross-multiply and solve:
30R = 15(30 + R),
30R = 450 + 15R,
30R - 15R = 450,
15R = 450,
R = 30\,\Omega.
Answer
\boxed{30\,\Omega}
Below is the originally provided solution reference:
