Answer:
I1 < I2
V1 > V2
Explanation:
Let the EMF of both the batteries is E and the internal resistance of both the cells is r.
The relation between the current and the EMF is given by
[tex]I = \frac{E}{R+r}[/tex]
where, R be the resistance connected in the circuit.
the relation between the terminal potential difference and the EMF is given by
V = E - Ir
where, V is the terminal potential difference
If R1 is connected,
The current is given by
[tex]I_{1} = \frac{E}{R_{1}+r}[/tex] .... (1)
The terminal potential difference is given by
[tex]V_{1}=E - I_{1}r[/tex] ..... (2)
If R2 is connected,
The current is given by
[tex]I_{2} = \frac{E}{R_{2}+r}[/tex] .... (3)
The terminal potential difference is given by
[tex]V_{2}=E - I_{2}r[/tex] ..... (4)
As given in the question, R2 > R1
So, by the equation (1) and (3), we get
I1 < I2
and by the equation (2) and (4) ,we get
V1 > V2