In the circuit shown in figure, the cell has emf $= 10 V$ and internal resistance $= 1 Ω$

Question

Q.

Circuit diagram shows that resistors $2 Ω$ , $4 Ω$ and $R Ω$ connected to a battery of e.m.f. 2 V and internal resistance $3 Ω$ . A main current of 0.25 A flows through the circuit. What is the p.d. across the $4 Ω$ resistor?

Answer 1

So current across

3

ohms will be

1 A

Answer 2

Since the current will be divided symmetrically in both the branches,
In first branch it will be halved is

0.5 A

After that it will be halved again i.e.

0.25 A

.
Thus the current across

4 o h m s

would be

0.25 A

In the circuit shown in figure, the cell has emf =10V and internal resistance =1Ω

So current across 3ohms will be 1A

Since the current will be divided symmetrically in both the branches,In first branch it will be halved is 0.5 AAfter that it will be halved again i.e. 0.25 A.Thus the current across 4 ohms would be 0.25 A

In the circuit shown in figure, the cell has emf $= 10 V$ and internal resistance $= 1 Ω$

So current across $3$ ohms will be $1 A$

Since the current will be divided symmetrically in both the branches,
In first branch it will be halved is $0.5 A$
After that it will be halved again i.e. $0.25 A$ .
Thus the current across $4 o h m s$ would be $0.25 A$