Answer:
The answer is "0.38 A".
Explanation:
Please find the missing image file in the attachment.
In the given image let:
[tex]R_{12} =R_1+R_2=16 \ \Omega \\\\R_{34} =R_3+R_4=20 \ \Omega \\[/tex]
[tex]R_P=\frac{R_{12} \cdot R_{34}}{R_{12} +R_{34}}[/tex]
[tex]= \frac{16 \cdot 20 }{16 +20}\\\\= \frac{320 }{36}\\\\= \frac{80 }{9}\\\\= 8.88 \ \Omega[/tex]
similarly,
[tex]R_{eq} =R_p+R_5+R_6\\\\[/tex]
[tex]=8.88+24+20\\\\=52.89 \ \Omega \\\\[/tex]
calculating the current value through a switch:
[tex]\to I=\frac{V}{R_{eq}} = \frac{20 \ v}{52.89 \Omega }=0.38\ A[/tex]
Answer:
D. 0.38 amperes
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
