Answer:
Explanation:
Given
[tex]R_1=10 \Omega [/tex]
[tex]R_2=5 \Omega [/tex]
when resistance in Parallel
[tex]\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]R_p=\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]R_p=\frac{10}{3}[/tex]
Suppose V is voltage of battery
Total Current [tex]i=\frac{3V}{10}[/tex]
Since Circuit is Parallel therefore Voltage across both resistor is same
[tex]V=i_1R_1=i_2R_2[/tex]
and [tex]i_1+i_2=i[/tex]
[tex]i_1+i_1\cdot \frac{R_1}{R_2}=i[/tex]
[tex]i_1(1+\frac{10}{5})=\frac{3V}{10}[/tex]
[tex]i_1=\frac{V}{10}[/tex]
[tex]i_2=\frac{2V}{10}[/tex]
(b) When Circuit is in series
[tex]R_s=R_1+R_2[/tex]
[tex]R_s=10+5=15 \Omega [/tex]
since circuit is in Series therefore current is same in both resistor
Current [tex]i=\frac{V}{15} A[/tex]
Voltage drop across [tex]R_1=i\times R_1[/tex]
[tex]V_1=\frac{V}{15}\times 10=\frac{2V}{3}[/tex]
[tex]V_2=\frac{V}{15}\times 5=\frac{V}{3}[/tex]