P 3.6-1 The circuit shown in Figure P 3.6-1a has been divided into two parts. In Figure

P 3.6-1b, the right-hand part has been replaced with an equivalent circuit. The left-hand part of the circuit has not been changed.

(a) Determine the value of the resistance R in Figure P 3.6-1b that makes the circuit in Figure P 3.6-1b equivalent to the circuit in Figure P 3.6-1a.

(b) Find the current i and the voltage v shown in Figure P 3.6-1b. Because of the equivalence, the current i and the voltage v shown in Figure P 3.6-1a are equal to the current i and the voltage v shown in Figure P 3.6-1b.

(c) Find the current i2 shown in Figure P 3.6-1a using current division.

 

Figure P 3.6-1

 

(a) Determine the value of the resistance R in Figure P 3.6-1b that makes the circuit in Figure P 3.6-1b equivalent to the circuit in Figure P 3.6-1a.

 

The equivalent resistor is the result of adding the 48 Ω and 24 Ω resistors in parallel

 

 

This equivalent resistance is then in series with the 16 Ω resistor so the value of R is

 

 

 

 

(b) Find the current i and the voltage v shown in Figure P 3.6-1b. Because of the equivalence, the current i and the voltage v shown in Figure P 3.6-1a are equal to the current i and the voltage v shown in Figure P 3.6-1b.

 

Since we have now have two 32 Ω resistors in parallel, the current i will be one half the total current.  We find the total current by finding the total resistance.  The two 32 Ω resistors in parallel can be replaced by a 16 Ω resistor and so the total resistance is

 

So total current is

 

So the current i shown is ½ A and we can voltage by doing a voltage drop

 

 

(c) Find the current i2 shown in Figure P 3.6-1a using current division.

 

So the 0.5 A current sees a 48 Ω resistor and 24 Ω resistor, so 2/3 of current will go through the 24 Ω resistor which is i2

 

 

 

 

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This page last updated on January 19, 2019