**Electric Current Exam1 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/R _{eq}=1/4+1/4**

**R _{eq}=2Ω**

Equivalent resistance between points C and B;

**R _{eq}=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 I_{1}, and current passing from 8Ω resistor is I_{2}, find I_{1}/I_{2}.

**a) **We redraw circuit and make it simple;

**R _{eq}=7+(2.2)/(2+2) (resistance in lower branch)**

**R _{eq}=8Ω**

**Equivalent resistance between points X and Y;**

**R _{eq}=(8.8)/(8+8)=4Ω**

**b) **

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

**V _{1}=I_{1}.R_{1} (lower branch)**

**V=I _{1}.8**

**V _{2}=I_{2}.R_{2} (upper branch)**

**V=I _{2}.8**

**I _{1}/I_{2}=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 opened;

Potential of B is equal to potential of D.

V_{A}>V_{B}=V_{D}

When we close switch;

potentials of the bulbs become;

**V _{A}=V**

**V _{B}=V_{C}=V/3**

**V _{D}=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**

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