Question

The circuit below is powered by a 20-volt battery. What current is flowing through the on/off switch (when it is on, as shown)?

Answer 1

D. 0.38 amperes

Step-by-step explanation:

Got it correct on Gizmos

Gizmos explanation: This compound circuit contains two parallel branches connected in series to 2 more resistors. Each parallel branch contains 2 resistors in series. The resistance of the first branch is 5+11=16 ohms, and the resistance of the second branch is 20 ohms. Using the equivalent resistance formula, the resistance of this section is

1R=116+120=580+480=980

So R is 809, or 8.89 ohms. This is connected in series to a 24-ohm and a 20-ohm resistor, for a total resistance of 52.89 ohms. The battery voltage is 20 volts, so by Ohm's law

I=VR

I=2052.89=0.38amperes

Answer 2

The answer is "0.38 A".

Step-by-step explanation:

Please find the missing image file in the attachment.

In the given image let:

`R_(12) =R_1+R_2=16 \ \Omega \\\\R_(34) =R_3+R_4=20 \ \Omega \\`

`R_P=(R_(12) \cdot R_(34))/(R_(12) +R_(34))`

`= (16 \cdot 20 )/(16 +20)\\\\= (320 )/(36)\\\\= (80 )/(9)\\\\= 8.88 \ \Omega`

similarly,

`R_(eq) =R_p+R_5+R_6\\\\`

`=8.88+24+20\\\\=52.89 \ \Omega \\\\`

calculating the current value through a switch:

`\to I=(V)/(R_(eq)) = (20 \ v)/(52.89 \Omega )=0.38\ A`