Question

Two cells one of emf 4V and internal resistance 0.2 ohm the other of emf 2.4 V and internal resistance 1 ohm are connected in parallel and the combination is connected in series with an external resistance of 10 ohm. Then a current which flows through the external resistance is

Answer 1

Answer:

ཐཅཆཅཆཐཕཧཝཝཕཐཐཧཐཕཕཐཕཕ

Explanation:

ཆཐཚཆཚཚཚཐཧཧཝཝཕཆཅཤཆཚཕཚཚཅྖཐཕཕཝཝཧཐཐཕཕཧཕཚཅཆཚཕཕཕཐཐཕཕཕཕཕཕཐ

ཚཐཕཕ

ཐཕཧཝཝཝཝཝཝཝཝཝཝཝཝཝ

ཧཧཝཝཝཝཝཝཕཐཆཆཅཅཨཤཤཐཧཝཞཝཕཆཕཚཚཚཐཕཝཝཝཝཝཝཝཝཝཇཐཅ

Answer 2

Given:

E1 = 4V

E2 = 2.4 V

r1 = 0.2 Ω

r2 = 1 Ω

R = 10 Ω

To Find:

The current which flows through the external resistance R.

Calculation:

- The equivalent emf will be:

E$_{eq}$ = (E1r2 + E2r1) / (r1 + r2)

⇒ E$_{eq}$ = (4 × 1 + 2.4 × 0.2) / (0.2 + 1)

⇒ E$_{eq}$ = 4.48 / 1.2

⇒ E$_{eq}$ = 3.73 V

- Current through external resistance R is given as:

I = E$_{eq}$/R

⇒ I = 3.73/ 10

⇒ I = 0.373 A

- So, the current which flows through the external resistance R is 0.373 A