1.751 V is the terminal voltage when a 7.00 Ω resistor is used
Answer: Option D
Explanation:
With [tex]15 \times 10^{8} \Omega[/tex] circuit resistance the current can be regarded as zero. So the potential difference (P.D) across the battery terminals is the emf (that means no P.D loss across the internal resistance). Hence,
Internal emf of battery = 2.50 V
Let we consider that the internal resistance = r Ω
P.D across 'r' as,
[tex]V_{r}=2.50-1.75=0.75 V[/tex]
Circuit current is given by,
[tex]I=\frac{\text {external } P . D}{5.00 \Omega}[/tex]
[tex]I=\frac{1.75}{5}=0.35 A[/tex]
Now, we can find as below with all data,
[tex]r=\frac{V_{r}}{I}=\frac{0.75}{0.35}=2.14 \Omega[/tex]
[tex]I=\frac{e m f}{r+R}=\frac{2.50}{5+2.14}=\frac{2.50}{7.14}=0.35 A[/tex]
[tex]V_{r}=I \times r=0.35 \times 2.14=0.749 V[/tex]
[tex]\text { Terminal } P . D\ or\ voltage =e m f - V_{r}=2.5-0.749=1.751 V[/tex]
