When R = 4 Ω and I = 1 A
We know that,
Terminal Voltage, V = E – Ir
So we have,
V = IR = 4 = E – Ir
or
E – r = 4 —– (1)
When R’ = 9 Ω and I’ = 0.5 A
V = IR = 0.5 x 9 = E – 0.5r
or
E – 0.5r = 4.5 —– (2)
By solving equation (1) and (2), we get,
r = 1 Ω and E = 5 V
Hence, the value of emf E = 5V and internal resistance r = 1Ω