Finding the supply voltage, knowing the resistance and current

Question

I am learning the basics of electrotechnics. I had this task for my university that I failed doing and I want to learn how to solve it anyway. Thanks for assistance! :)

Here is the exercise task: I assume I need to find the \$U\$. Knowing the \$I = 14 A\$ and \$R = 8 Ω\$

This is how I calculated this:

\$ U = IR_{z} \$ where \$ R_{z} = R_{123456}\$ (equivalent resistance)

\$ R_{12} = R_{1} + R_{2} = 4 Ω + 4 Ω = 8 Ω \$

\$ R_{123} = \frac{R_{12}R_{3}}{R_{12}+R_{3}} = \frac{8 Ω * 8 Ω}{8 Ω + 8 Ω} = 4 Ω\$

\$ R_{1234} = R_{123} + R_{4} = 4 Ω + 4 Ω = 8 Ω \$

\$ R_{12345} = \frac{R_{1234}R_{5}}{R_{1234}+R_{5}} = \frac{8 Ω * 8 Ω}{8 Ω + 8 Ω} = 4 Ω\$

\$ R_{123456} = R_{12345} + R_{6} = 4 Ω + 4 Ω = 8 Ω \$

\$ R_{z} = R_{123456} = 8 Ω \$

\$ U = IR_{z} = 14 A * 8 Ω = 112 V \$

Yet somehow 112 V is wrong answer. Any clues, hints? Where did I do an error?

Answer 1

You calculated the voltage so that the total current provided by the supply is 14 A. But not all of that current goes through the branch identified by the "I" in the schematic. So now you need to figure out what fraction of the total current is I, and scale your input voltage appropriately to get 14 A at "I".