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Question: 10V 1$\Omega$ 2$\Omega$ A B E₁ r=1$\Omega$ G K₁ R...

10V 1Ω\Omega

2Ω\Omega

A B

E₁ r=1Ω\Omega G

K₁

R

Answer

No specific question asked.

Explanation

Solution

The circuit is a potentiometer.

  1. Primary circuit: Calculate the potential difference across the potentiometer wire AB. VAB=10V1Ω+2Ω×2Ω=203VV_{AB} = \frac{10V}{1\Omega + 2\Omega} \times 2\Omega = \frac{20}{3} V.
  2. Potential gradient: k=VABL=203LV/mk = \frac{V_{AB}}{L} = \frac{20}{3L} V/m, where L is the length of the wire AB (not given).
  3. Secondary circuit (K₁ open): The galvanometer balances the EMF E₁. At balance, E1=kL1E_1 = k L_1, where L1L_1 is the balancing length.
  4. Secondary circuit (K₁ closed): The resistor R is connected. Assuming R shunts E₁, the terminal voltage VtV_t across E₁ is balanced. At balance, Vt=kL2V_t = k L_2, where L2L_2 is the balancing length.
  5. Internal resistance: r=R(L1L2−1)r = R \left( \frac{L_1}{L_2} - 1 \right).

Without a specific question or numerical values for lengths/resistance R, a definitive numerical answer cannot be provided. The circuit is used for measuring EMF and internal resistance of a cell.

Answer: No specific question was asked in the prompt. The provided image is a potentiometer circuit typically used for measuring the EMF of a cell and its internal resistance.