Answer :
- Total resistance is 10Ω
- Current through the ammeter is of 0.9 A
Explaination :
As we know that, if n resistances R1 , R2 and R3 .... Rn are joined in parallel the equivalent resistance is given as :
[tex] \boxed{ \sf{ \dfrac{1}{R_p} \: = \: \dfrac{1}{R_1}\: + \: \dfrac{1}{R_2} \: + \: \dfrac{1}{R_3} \: \: .... \: \dfrac{1}{R_n} }}[/tex]
R1 = 20Ω
R2 = 20Ω
Putting the values,
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \dfrac{1}{20Ω}\: + \: \dfrac{1}{20Ω}}[/tex]
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \dfrac{1 \times 1}{20}\: + \: \dfrac{1 \times 1}{20}}[/tex]
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \: \dfrac{1 \: + \: 1}{20}}[/tex]
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \: \dfrac{2}{20}}[/tex]
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \: \cancel\dfrac{2}{20}}[/tex]
[tex]: \longmapsto \: \sf{ \dfrac{1}{R_p} \: = \: \dfrac{1}{10}}[/tex]
[tex]: \longmapsto \: \pink{\bf{R_p\: = \: 10Ω }}[/tex]
Current through the ammete:-
As we know that,
- [tex]\boxed{ \sf{I \: = \: \frac{V}{R } }}[/tex]
Here ,
- I is current
- V is Volts
- R is resistance
Putting the values,
[tex]\longmapsto \: \sf{I \: = \: \dfrac{9}{10 } }[/tex]
[tex] \longmapsto \: \red{\bf{I \: = \: 0.9}}[/tex]
