Parallel and Series Circuits

Question 1 

The relative brightness of the bulbs in the circuit will be ranked as follows: 

A>D>B=C 

Ideally, the brightness of a bulb depends on the amount of current. Bulb A will be the brightest in the system because it has the largest voltage drop compared to other bulbs, meaning that it has the highest amount of current flowing through it. In addition, the total amount of current is constant throughout the electric system ( Qawasmi, n.d. ). Therefore, the initial current, I, leaving the positive terminal of the battery passes through A. However, due to the parallel arrangement of bulbs D and B+C, the initial current splits into two. The initial current will be divided based on the resistance of the bulbs. Since D is less resistive compared to bulbs B+C, higher current will tend to flow through bulb D compared to bulbs B+C, making bulb D brighter compared bulbs B and C. In other words, if the circuit had an initial current, I, bulb D will get more than I/2 while bulbs B+C will get less than I/2. In addition to being the dimmest in the circuit, bulbs B and C have identical levels of brightness ( Qawasmi, n.d. ). Since the bulbs B and C are identical to each other, and are in a series arrangement to each other, they will posses the same voltage drop, and thus the same brightness. 

Question 2 

The relative brightness of the bulbs in the circuit will be ranked as follows: 

C>A>D>B 

As noted earlier, the amount of current is constant throughout the circuit. However, at the junction between the ammeter and bulb A, the current splits depending on the resistance on each arrangement. Since an ideal ammeter has a relatively lower resistance, it will attract greater amount of current compared to the current going through bulb A ( Qawasmi, n.d. ). The current flowing through the ammeter, in addition to current that went through bulb B will go through bulb C. Resultantly, bulb C has the highest current flowing through it, and thus the relatively higher level of brightness compared to other bulbs. Bulb A ranks second because it has the second highest amount of current going through it. The initial current is split between the ammeter and bulb A junction. The remainder of the initial current goes through bulb A. However, the subsequent bulbs, B and D, share the remainder of the initial current among themselves. Therefore, bulb A will automatically be brighter than bulbs B and D. Elsewhere, since the resistance across bulb B and C will be greater than the resistance in bulb D, greater current will be drawn towards bulb D. Therefore, bulb D will be brighter compared to bulb B. 

When compared to the circuit in question 1, the brightness of bulb D decreases. The current flowing through bulb D in the circuit in question 1 is higher compared to the one flowing in the circuit in question 2. The addition of the ammeter into the circuit creates a short route that diverts a significant amount of current; the current through the ammeter does not travel across bulb D. Besides, unlike in the initial circuit, where more than half of the initial current travelled across bulb D, bulb D shared the less than half of the initial current with bulbs B and C. 

Question 3 

The relative brightness of the bulbs in the circuit will be ranked as follows: 

A>D>B=C 

The current going through bulb A is greater than the current going through the other bulbs. The whole of the initial current coming from the battery flows through bulb A, making it the brightest. However, when the initial current reaches the junction, it is shared across the two parallel paths depending on the level of resistance on the paths ( Qawasmi, n.d. ). Resultantly, bulbs, B, C, and D, automatically become dimmer compared to A. Of the remaining three bulbs, bulbs D is the brightest. The current in the parallel path with bulb D is larger compared to that in the route with bulbs B and C; the parallel path with bulbs B and C has a greater load and thus attracting less current to flow through it ( Qawasmi, n.d. ). Bulb B and C have the same current flowing through them, and thus possessing same level of brightness. Since an ideal voltmeter does not posses great resistance, its impact on current distribution across the circuit is insignificant. 

The reading of the voltmeter is V=0. There are no voltage drawing elements after bulb C all the way to the negative side of the batteries. The potential difference across the circuit should be V bat . Therefore, while the voltage just before bulb A will read V= V bat , the cumulative voltage drop between bulb A and bulb C should be equal to V bat . Furthermore, in order for the potential difference of V bat to be maintained across the battery, the voltage at the location of the voltmeter should be zero. 

The introduction of the voltmeter has not hugely affected the distribution of current across the parallel paths in the circuit. The ideal voltmeter does not increase the load in the parallel path with bulbs B and C ( Qawasmi, n.d. ). Therefore, the brightness of bulb D remains relatively the same compared to the circuit in question 1. 

Question 4 

Part (a) 

1.  

Based on the arrangement, the two batteries on the left and right produces clockwise and counter-clockwise currents, respectively. Therefore, since the batteries are identical, the cumulative current flowing through the circuit is zero. With no current flowing through the system, no power is dissipated in the resistors. 

1.  

Assuming current from the two batteries being I 1 and I 2 , Kirchhoff’s laws can be used: 

+V o1 – 2RI 1 + V o2 =0 

since V o1 and V o2 are identical: 

2V o = 2RI 1 

I 1 = 

+V o2 – RI 2 =0 

I 2 = 

Therefore, 

Power through resistor (2R), P = I 2 R = (  2 ×2R = 

Power through resistor (R), P = I 2 R = (  2 ×R= 

1.  

V o1 -R (I 1 +I 3 ) = 0 (equation 1) 

V o2 -2R (I 4 ) = 0 (equation 2) 

I 3 + I 4 = I 2 

V o1 = R (I 1 +I 3 ) 

(I 1 +I 3 ) = 

I 4 = 

Power through resistor (R), P = I 2 R = (  2 × R = 

Power through resistor (R), P = I 2 R = (  ) 2 × 2R = 

1.  

V o – 2RI 1 +V o – RI 2 = 0 

I=  

since I = I 1 + I 2 , 

I 1 =  – I 2 

2V o = 2R (  – I 2 ) +RI 2 

2V o =  -2RI 2 +RI 2 

RI 2 = - 

I 2 = - 

Power through resistor (R), P = I 2 R = (  2 R = - 

Power through resistor (2R), P = I 2 R = (  – I 2 ) 2 2R = (  +  ) 2 2R = 

Circuit 1 will have the longest lifetime because no voltage is lost. Since the resultant current is zero, there will be no voltage drop at any of the resistors. Circuit 2 will have the second longest lifetime because the first battery to die will be due to only an R resistor. The two resistors in Circuit 4 shares the two batteries, meaning its rate of depletion will be relatively lower compared to Circuit 3, which has a 2R resistor relying on a single battery. 

Question 5 

V = IR 

 = I 1 ×5R 

I 1 = 

According to Kirchhoff’s law, the amount of current entering a node is equal to the amount of current leaving a node. Therefore: 

I bat = I 1 + I 3 

Since I 1 is known, I 3 will be: 

I 3 = I bat – I 1 = I bat - 

Since there is no any other junction between R 1 and R 2 , the current across the two resistors will remain the same. Therefore: 

I 1 = I 2 = 

Since the current leaving a node should be equal to the current entering a node, the current flowing through R 4 will be: 

I 4 = I 1 + I 3 =  I bat -   = I bat 

Reference 

Qawasmi, E. (n.d.) Ohm's and Kirshoff's Law (Part One). Khalifa University.