What are the (a) size and (b) direction (up or down) of current $i$in Fig. 27-71, where all resistances are$4.0\Omega$and all batteries are ideal and have an emf of 10V? (Hint: This can be answered using only mental calculation.)

• a)Resistance of all the given resistors, R=4.0$\Omega$
• b) Emf of an ideal battery,$\epsilon =10V$

We know that the current flows through the positive to the negative terminal of the battery. Thus, a battery placed in a reverse way acts as a resistor with a high resistance value and hence opposes the current flow. Again for the given case of the resistors and batteries with reverse to the current flow blocks the current while others play the role in current flow. Thus, using this concept, the equivalent resistance and the net emf of the entire circuit are calculated that are further used to calculate the size and direction of the current.

Formula:

The voltage equation using Ohm’s law,$V=IR$

The equivalent resistance for a series combination,$R_{eq}=\underset{1}{\overset{n}{\sum R_n}}$ (ii)

The equivalent resistance for a parallel combination,$R_{eq}=\underset{1}{\overset{n}{\sum \frac{1}{R_n}}}$ (iii)

(a)

The resistor by the letteriis above three other resistors.

Now, the equivalent resistance of these four resistors can be given using equations (ii) and (iii) as follows:

$$\begin{gathered} R_{eq}=R+\frac{R^2}{2R}+R \\ =5RI2 \\ =5(4.0\Omega )/2 \\ =10.0\Omega \end{gathered}$$

The current flowing through this equivalent resistance is given by i as the current value remains unchanged for series resistors.

As per the concept, we can see that the current through other resistors differ and also oppose the current flow, thus, we can neglect them as we need the value of current i only.

Now, we can consider the emfs that follow the similar arrangement pattern as that to the battery connected directly to the equivalent resistance and allow the current flow, this will given us the number of operating emf batteries in the given maze as 4.

Thus, the net emf becomes:
$$\begin{gathered} \epsilon \text{'}=4\left(10V\right) \\ =40V \end{gathered}$$

Thus, the current value through the circuit is given by using equation (i) as follows:

$$\begin{gathered} i=\frac{40V}{10.0\Omega \\ } \\ =4.0A \end{gathered}$$

Hence, the size of the current is 4.0A.

(b)

The direction of the current flowing for the found equivalent resistance is upward as the battery attached to the resistance passes the current in the direction from down to up that is from positive to negative terminal of the battery in the figure.

Hence, the direction of current is upward.