Battery Life worksheet

1. Estimated runtime: 6.29 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{2,750}\text{\hspace{0.17em}mAh}\times \mathrm{64\%}$$=\mathrm{1,760}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{1,760}\text{\hspace{0.17em}mAh}}{280\text{\hspace{0.17em}mA}}$$=6.285714\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $6.2857\text{\hspace{0.17em}h}$$=\text{6 h 17 min 9 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{2,750}\times 4}{\mathrm{1,000}}$$=11\text{\hspace{0.17em}Wh}$

2. Estimated runtime: 9.26 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{2,500}\text{\hspace{0.17em}mAh}\times \mathrm{100\%}$$=\mathrm{2,500}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,500}\text{\hspace{0.17em}mAh}}{270\text{\hspace{0.17em}mA}}$$=9.259259\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $9.2593\text{\hspace{0.17em}h}$$=\text{9 h 15 min 33 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{2,500}\times 2.4}{\mathrm{1,000}}$$=6\text{\hspace{0.17em}Wh}$

3. Estimated runtime: 12.25 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{3,500}\text{\hspace{0.17em}mAh}\times \mathrm{70\%}$$=\mathrm{2,450}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,450}\text{\hspace{0.17em}mAh}}{200\text{\hspace{0.17em}mA}}$$=12.25\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $12.25\text{\hspace{0.17em}h}$$=\text{12 h 15 min}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{3,500}\times 3.2}{\mathrm{1,000}}$$=11.2\text{\hspace{0.17em}Wh}$

4. Estimated runtime: 9.1 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{3,250}\text{\hspace{0.17em}mAh}\times \mathrm{98\%}$$=\mathrm{3,185}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{3,185}\text{\hspace{0.17em}mAh}}{350\text{\hspace{0.17em}mA}}$$=9.1\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $9.1\text{\hspace{0.17em}h}$$=\text{9 h 6 min}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{3,250}\times 3.6}{\mathrm{1,000}}$$=11.7\text{\hspace{0.17em}Wh}$

5. Estimated runtime: 7.9 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{3,600}\text{\hspace{0.17em}mAh}\times \mathrm{68\%}$$=\mathrm{2,448}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,448}\text{\hspace{0.17em}mAh}}{310\text{\hspace{0.17em}mA}}$$=7.896774\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $7.8968\text{\hspace{0.17em}h}$$=\text{7 h 53 min 48 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{3,600}\times 4}{\mathrm{1,000}}$$=14.4\text{\hspace{0.17em}Wh}$

6. Estimated runtime: 10.88 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{3,200}\text{\hspace{0.17em}mAh}\times \mathrm{85\%}$$=\mathrm{2,720}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,720}\text{\hspace{0.17em}mAh}}{250\text{\hspace{0.17em}mA}}$$=10.88\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $10.88\text{\hspace{0.17em}h}$$=\text{10 h 52 min 48 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{3,200}\times 4.4}{\mathrm{1,000}}$$=14.08\text{\hspace{0.17em}Wh}$

7. Estimated runtime: 8.44 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{2,000}\text{\hspace{0.17em}mAh}\times \mathrm{95\%}$$=\mathrm{1,900}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{1,900}\text{\hspace{0.17em}mAh}}{225\text{\hspace{0.17em}mA}}$$=8.444444\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $8.4444\text{\hspace{0.17em}h}$$=\text{8 h 26 min 40 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{2,000}\times 3.4}{\mathrm{1,000}}$$=6.8\text{\hspace{0.17em}Wh}$

8. Estimated runtime: 7.27 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{2,750}\text{\hspace{0.17em}mAh}\times \mathrm{74\%}$$=\mathrm{2,035}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,035}\text{\hspace{0.17em}mAh}}{280\text{\hspace{0.17em}mA}}$$=7.267857\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $7.2679\text{\hspace{0.17em}h}$$=\text{7 h 16 min 4 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{2,750}\times 3.4}{\mathrm{1,000}}$$=9.35\text{\hspace{0.17em}Wh}$

9. Estimated runtime: 17.5 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{3,750}\text{\hspace{0.17em}mAh}\times \mathrm{84\%}$$=\mathrm{3,150}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{3,150}\text{\hspace{0.17em}mAh}}{180\text{\hspace{0.17em}mA}}$$=17.5\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $17.5\text{\hspace{0.17em}h}$$=\text{17 h 30 min}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{3,750}\times 3.8}{\mathrm{1,000}}$$=14.25\text{\hspace{0.17em}Wh}$

10. Estimated runtime: 9.35 h

1. Keep only the usable share of the capacity, since some is lost in real use

   $\mathrm{2,700}\text{\hspace{0.17em}mAh}\times \mathrm{90\%}$$=\mathrm{2,430}\text{\hspace{0.17em}mAh}$

2. Divide the usable charge by the current the device draws. This gives the runtime in hours

   $t$$=\frac{\mathrm{2,430}\text{\hspace{0.17em}mAh}}{260\text{\hspace{0.17em}mA}}$$=9.346154\text{\hspace{0.17em}h}$

3. Write the runtime in days, hours and minutes

   $9.3462\text{\hspace{0.17em}h}$$=\text{9 h 20 min 46 s}$

4. For reference, the capacity in watt-hours at this voltage: Wh = mAh × V ÷ 1,000

   $\frac{\mathrm{2,700}\times 3.4}{\mathrm{1,000}}$$=9.18\text{\hspace{0.17em}Wh}$