# Electric Current Exam 1 and Problem Solutions

**1.** Voltage vs. current graph of a conductor is given below. Find the change in the resistance of conductor in first and third intervals.
We use ohm’s law to find relation between V, I and R.

**Interval I:** Since potential and current increase linearly, resistance of the conductor becomes constant.

**Interval II:** In this interval, potential is constant but current increases. Thus, resistance of the conductor must decrease to make potential constant.

**Interval III:** In this interval potential increases but current is constant. Thus, resistance of the conductor must increase to make potential increase.

**2. a)** Find the equivalent resistance of the circuit given below.

**b)** Find current passing through the circuit.
**a)** Since resistance are connected in parallel between points A and B;
**1/Req=1/4+1/4**

**Req=2Ω**

Equivalent resistance between points C and B;
**Req=4+2=6Ω**

**b) V=I.R**

**36=I.6**

**I=6Amperes**

**3.** **a)** Find equivalent resistance between the points X and Y.

**b)** If the current passing from 7Ω resistor is I₁, and current passing from 8Ω resistor is I₂, find I₁/I₂.
**a)** We redraw circuit and make it simple;
**Req=7+(2.2)/(2+2)** (resistance in lower branch)

**Req=8Ω**

Equivalent resistance between points X and Y;

**Req=(8.8)/(8+8)=4Ω**

**b)**
Since resistances are in parallel, their potentials are equal. Using ohm’s law;

**V₁=I₁.R₁** (lower branch)

**V=I₁.8**

**V₂=I₂.R₂** (upper branch)

**V=I₂.8**

**I₁/I₂=1**

**4.** If we close switch shown in the picture given below, find the changes in the brightness of the bulbs.
Circuit when switch is closed;
Potential of B is equal to potential of D.

**VA>VB=VD**

When we close switch; potentials of the bulbs become;

**VA=V**

**VB=VC=V/3**

**VD=2V/3**

Thus; there is no change in the brightness of the bulb A, brightness of B decreases, and D increases.

**5.** Find the efficiency of the motor in the circuit given below.
Current passing through circuit is;

**I=(ε₁+ε₂-ε’)/Req**

**I=(40+60-70)/(7+2+2+4)**

**I=2 Amperes**

**Efficiency=ε’/(ε’+I.r’)=70/(70+2.2)**

**Efficiency=35/37**