Inspired by daily challenges #2

In the previous problem, none of the bulbs were in danger of burning out, and the introduction of resistance makes them even less likely to burn out in this case.

In Setup B, the 5 ohms of series bulb resistance still takes most of the total voltage since it is larger than the 2 ohm switch resistance. So Setup B still functions similarly to before, but with the bulbs a bit dimmer.

In the previous problem, each bulb in Setup A had twice the battery voltage across it. Now, each bulb in Setup A has dramatically less voltage, due to the voltage divider between the 2 ohm switch resistance and the equivalent 0.2 ohm parallel set of bulbs.

So the brightness of Setup A is dramatically reduced relative to the previous problem, while the brightness of Setup B is only moderately reduced.

A nice quantitative followup problem to this would be to ask for the ratio of the Setup A total bulb power to the Setup B total bulb power.

Let the voltage of a D-cell battery be $V \text{ V}$ . Since the resistance of a bulb is $1 \ \Omega$ Then the rated current of each bulb is $\dfrac {2V}{1} = 2V \text{ A}$ and the bulb would burn out if the current through it is $I_b = 3 V \text{ A}$ .

The equivalent circuit of the two setup is as shown above.

Setup A: The 5 $\times$ 2 D-cell batteries in parallel give a $V_b = 2V \text{ V}$ . The 2 $\times$ 100-m wires in series make $R_w = 2 \ \Omega$ . The 5 bulbs in parallel make $R_b = \dfrac 15 = 0.2 \ \Omega$ . The current through the circuit $I = \dfrac {V_b}{R_w+R_b} = \dfrac {2V}{2.2} = \dfrac {10}{11}V$ . The current through each bulb $I_b = \dfrac I5 = \dfrac 2{11}V < 3V$ . So the bulb would not burn out.

Setup B: For 10 D-cell batteries connected in series, $V_b = 10V \text{ V}$ . The 2 $\times$ 100-m wires in series make $R_w = 2 \ \Omega$ . The 5 bulbs in series make $R_b = 5 \ \Omega$ . The current through the circuit $I = \dfrac {V_b}{R_w+R_b} = \dfrac {10}7 V$ . The current through each bulb $I_b = I = \dfrac {10}7V < 3V$ . So the bulb would not burn out.

We note that the 5 bulbs in Setup B have a higher current than the 5 in Setup A, therefore "B" produces more light .

As indicated in the diagram.

Batteries are connected in five sets of in two cell series

Example battery A and B are in series. Therefore, the red rail will be 1.5 +1.5 = 3 volts above the green rail. The battery supply voltage is 3.0volts.

The five lamps are in a parallel arrangement and are in series with two wire resistances. Calculating the resulting resistance of the lamp/wire arrangement. $\frac { 1 }{ { R }_{ A } } +\frac { 1 }{ { R }_{ B } } +\frac { 1 }{ { R }_{ C } } +\frac { 1 }{ { R }_{ D } } +\frac { 1 }{ { R }_{ E } } = \frac { 1 }{ { R }_{ lamps } } \\ \\ { R }_{ lamps }\quad = \frac { 1 }{ 5 } =\quad 0.2 ohms \\ { R }_{ total } = { R }_{ lamps } +{ R }_{ wire 1 }+ { R }_{ wire 2 }\\ { R }_{ total\\ } = 0.2 + 1 + 1 = 2.2 ohms$

calculate the total current that the batteries are supplying.

$I =\frac { V }{ R } = \frac { 3 }{ 2.2 } = 1.36 amps$

calculate voltage drop in the long small gauge wires $V=IR\quad$ calculate the voltage that is across the lamps after the line loss do to very high wire resistance to the switch

${ V }_{ lamps }=V_{ total \\ }-{ V }_{ wire }-{ V }_{ wire\\ }\\ { V }_{ lamps }=3-1.36-1.36=0.28volts$

clearly this is not a good circuit to control the bulbs for 100 meters with this small gauge wires. Total power used by each bulb can be calculated to show how little power is used to possibly make light.

${ W }_{ lamp }={ V }^{ 2 }/R=\frac { { 0.28 }^{ 2 } }{ 1 } =0.078 Watts/bulb$

There would be no light at all from any of the lamps, most of the battery energy is going to heat the long switch wires.

As indicated in the diagram.

All ten Batteries are connected in a series circuit. Each battery in the series will add 1.5 volts to the output voltage. 1.5v X 10= 15 volts output from the series of batteries.

All five of the bulbs and the two connecting wire are in a series circuit. calculate the total resistance of that circuit.

${ R }_{ total }={ R }_{ lamps }+{ R }_{ wires }=5+1+1=7ohms$

calculate current supplied by batteries.

$I=\frac { V }{ R } =\frac { 15 }{ 7 } =2.14amps$

Calculate the voltage drop across each of the bulbs. by applying ohms law. Find the the lamp voltage drop using the current through lamp, the lamps resistance
${ V }_{ drop\\ \\ \\ \\ }=2.14A\ast 1\Omega =2.14v$

calculate the total power used by each bulb to determine how much energy there may be to produce light.

${ W }_{ lamp }={ V }^{ 2 }/R=\frac { 2.14^{ 2 } }{ 1 } =4.58Watts/bulb$

much more of the total power produced by the batteries is delivered to the lamps . But still not good.

calculate the energy the each lamp would receive if the long switch wires where not in use.

${ W }_{ lamp }={ V }^{ 2 }/R\frac { { 3A }^{ 2 } }{ 1\Omega } =9.0Watts/bulb$

~ 50% less power per bulb do to line loss on his switch wires, perhaps Georg should consider using larger gauge wires to his switch.