Question

Find the current flowing out of the battery in the circuit. I = [?] A 9.0 V 30 ww 40 Ω www 50 Ω 20Ω 10Ω​

Answer 1

We need Net resistance

solve from right words

`\\ \rm\Rrightarrow R_1=20+10=30\Omega`

`\\ \rm\Rrightarrow (1)/(R_2)=(1)/(30)+(1)/(50)=(5+3)/(150)`

`\\ \rm\Rrightarrow R_2=(150)/(8)=18.75\Omega`

`\\ \rm\Rrightarrow (1)/(R_3)=(1)/(30)+(1)/(40)=(4+3)/(120)`

`\\ \rm\Rrightarrow R_3=(120)/(7)=17.14\Omega`

`\\ \rm\Rrightarrow R_(net)=17.14+18.75=35.89\Omega`

Use ohm's law

`\\ \rm\Rrightarrow I=(V)/(R)`

`\\ \rm\Rrightarrow I=(9)/(35.89)`

`\\ \rm\Rrightarrow I\approx 0.25A`