Share With

Four resistors are connected to a battery as shown in the figure below.

(a) Determine the potential difference across each resistor in terms of e m f . (Use the following as necessary: script E for e m f .)

(b) Determine the current in each resistor in terms of I. (Use the following as necessary: I.)

pur-new-sol

First we get the current in therm of the emf.

we get the equivalent resistance:

R2 and R3 are in series we get:

$R_A = R_2+R_3 = 6R$

then RA and R4 are in parallel so we get.

$R_B = \frac{R_AR_4}{R_A+R_4}=\frac{6R(3R)}{6R+3R}=2R$

Then RB and R1 are in series:

$R_{eq} = R_B+R_1 =2R+R=3R$

Using Ohm's law we get:

$V = I R_{eq}$

$\varepsilon = I (3R)$

so:

$I = \frac{\varepsilon}{3R}$

The current in RA and R4 following:

$2I_A = I_4$

and:

$I_4+I_A = I$

$I_A = \frac{I}{3}=\frac{\varepsilon}{9R}$

$I_4 = \frac{I}{3}=\frac{2\varepsilon}{9R}$

a) Using ohm's law for each resistance:

$\Delta V_1 = IR_1 =\frac{\varepsilon}{3R}R =\frac{\varepsilon}{3}$

$\Delta V_2 = I_AR_2 =\frac{\varepsilon}{9R}2R =\frac{2\varepsilon}{9}$

$\Delta V_3= I_AR_3 =\frac{\varepsilon}{9R}4R =\frac{4\varepsilon}{9}$

$\Delta V_4= I_4R_4 =\frac{2\varepsilon}{9R}3R =\frac{2\varepsilon}{3}$

b) For current:

$I_1 = I$

$I_2 = I/3$

$I_3 = I/3$

$I_4 = 2I/3$