Question

A Science Fiction Story Problem: Many Years in the future (2166 A.D.) the Mexican army builds a great wall (see next page for construction details) to keep the hordes of Merkn's to the north from illegally crossing the border (They want cross in order to escape persecution and torture at the hands of the totalitarian fascist government that has seized control up there). The wall is made of bricks that are 3.000"deep x 5.000"wide x 2.000" tall, held in place by an exactly " thick cement mortar (which implies that 1/8" of mortar edging is associated with each brick). The interior of the wall is a 2.71000 foot wide dirt core. (see figure). The air on the Merke side of the wall is at a temp of 1234.5 C due to the napalm bombing of El-Paso, while the air on the Juarez Mexico side of the wall is at a temp of 1.3 C because of the nuclear winter triggered by the "mistaken" war that the US pre-emptively launched against Canada those many years ago. Thermal Conductivity of brick =0.600 W/(mK) - Thermal Conductivity of dirt -1,50 W/(mK) - Thermal Conductivity of mortar = 1.732 W/(mK) - Heat transfer coefficient (h) for convection on BOTH sides of wall - 15.000 (W/m K). 5.000" Repeat Pattern 2.000" Brick & Mortar Wall surface 10.125" mortar border 3.00" deep brick TI 2.71 foot Convection ---- 0302 Dirt Core Convection LEBA MW

Answer 1

We will agree with Arthur. As there is only one path, so the rate of flow of charge i.e. current flowing through the circuit should be same every where. Consider any point in the circuit, if any electron comes towards that, same electron should leave as well, there cannot be any accumulation of electrons at any point, it is in accordance with the law of conservation of charge. As the bulbs C and B are brighter than A and D, their ohmic resistance must be more than that of A and D. If I is the current in the circuit, the power dissipated in each bulb is given by the formula, P = I^2*R. So higher the resistance, more the power dissipated in that bulb and higher will be its brightness as the current is same.

Let us suppose 0.1 A is the current flowing in the circuit. Resistance of B and C is 2$\Omega$ each and that of A and D 1$\Omega$ each. Then power dissipated in both B and C is P1=(0.1)^2*2 watt = 0.02 W each and in both A and D it is P2 = (0.1)^2*1 W=0.01 W each. Therefore we can see that P1 > P2 and hence B and C will glow brighter.

Potential differences across both A and D are same as potential difference is given by the relation, V = I*R, where R is the resistance and I is the current. As both current and resistance are same for both A and D, hence there is same potential difference across each one of them.