In the circuit shown in Fig. E26.17,the voltage across the$\mathbf{2}\mathbf{\Omega }$resistor is 12 V. What are the emf of the battery and the current through the$\mathbf{6}\mathbf{\Omega }$resistor?

Ohm's law states that thecurrentthrough aconductorbetween two points is directlyproportionalto thevoltageacross the two points. Introducing the constant of proportionality, theresistance, one arrives at the usual mathematical equation that describes this relationship:

Given
Let us label the next on the

$\mathrm{R}1=1.0\mathrm{\Omega R}2=2.0\mathrm{\Omega }\mathrm{and}\mathrm{R}3=6.0\mathrm{\Omega }$

The voltage across R2 is Vg = 12.0 V

Solution

thecurrentI3andtheemfofthebattery8-ThetworesistorsR1andR2areinseriesandthecurrent

through these two resistors is the same through every resistor and equals the current through the combination, and We could

calculate this current using Ohm's law

$$\begin{gathered} {\mathrm{R}}_1\mathrm{and}{\mathrm{R}}_2 \\ {\mathrm{l}}_{12}={\mathrm{l}}_1={\mathrm{II}}_2=\frac{{\mathrm{V}}_2}{{\mathrm{R}}_2} \\ \frac{12}{2}=6.0\mathrm{A} \end{gathered}$$

AsshOWnbythe?gure,thecombinationR12isinparallelwithR3andthepotentialdifferenceVacrossresistorsconnected
in the parallel is the same for every resistor and equals the potential difference across the combination as next
${\mathrm{V}}_{12}={\mathrm{V}}_3={\mathrm{V}}_{123}$
The combination voltage ${\mathrm{V}}_{123}$isthevoltagedropofthebatteryand$\hat{\mathrm{i}}={\mathrm{v}}_{123}$ as the internal resistance of the battery is zero,therefore the emf of the battery equals where We can get the combination voltage Vlz by Ohm's law in the next form
m we parallel IS me same nor every resuswr ano equals me pomenual omerence across me comomauon as The combination voltage V123 is the voltage drop of the

battery and as the internal resistance of the battery is zero,

therefore the emf of the battery equals${\mathrm{V}}_{12}-{\mathrm{V}}_3={\mathrm{V}}_{123}$where We can get the combination voltage V12 by Ohm's law in the

next form

$$\begin{gathered} {\mathrm{V}}_{12}-{\mathrm{V}}_3={\mathrm{V}}_{123} \\ \hat{\mathrm{i}}={\mathrm{v}}_{123}={\mathrm{I}}_{12}({\mathrm{R}}_1+{\mathrm{R}}_2) \\ =6(1+2) \\ =18\mathrm{V} \end{gathered}$$

Therefore the EMF of the battery is 18.00V

Now We can use the value of V to get the currentI${}_3$ where we can plug our values for V3 and R3 into Ohm's law to get I${}_3$

$$\begin{gathered} {\mathrm{I}}_3=\frac{{\mathrm{V}}_3}{{\mathrm{R}}_3} \\ =\frac{18}{6}=3\mathrm{A} \end{gathered}$$

The Current in 6 ohms resistor is 3A