Since the internal resistance is small, the current through the circuit will be large, I=εR+r=ε0+r=εr.The large current causes a high power to be dissipate...Since the internal resistance is small, the current through the circuit will be large, I=εR+r=ε0+r=εr.The large current causes a high power to be dissipated by the internal resistance (P=I2r).
3. P=I2R=(εr+R)2R=ε2R(r+R)−2,dPdR=ε2[(r+R)−2−2R(r+R)−3]=0[(r+R)−2R(r+R)3]=0,r=R In parallel, the terminal voltage is the s...3. P=I2R=(εr+R)2R=ε2R(r+R)−2,dPdR=ε2[(r+R)−2−2R(r+R)−3]=0[(r+R)−2R(r+R)3]=0,r=R In parallel, the terminal voltage is the same, but the equivalent internal resistance is smaller than the smallest individual internal resistance and a higher current can be provided. The ammeter has a small resistance; therefore, a large current will be produced and could damage the meter and/or overheat the battery.
